概率DP入门小结

说是概率DP，其实主要是求概率和期望的问题

说到DP总要有状态，每种状态可能有多种子状态

一般的DP是这样：在DP过程中，当前状态必然是由多个子状态中的最优的转移而来

所以一般的DP求的是最优的结果

而概率不需要最优，而是实际概率

所以概率DP最大的区别在于：在DP过程中，当前状态是由所有子状态的概率共同转移而来

所以概率DP只是利用了DP的动态而没有规划（只有状态转移，而不需要进行决策）

至于状态转移方程怎么列，最科学的自然是根据数学知识列，

不过实际做题中会发现找规律也是一种不错的方法，

而事实证明，如果可以状态转移，找规律的方法往往是可行的

不过数学扎实的话用数学知识绝对要比找规律快且准

POJ 3744 (矩阵优化)

题意：一条路上有n个地雷，你站在起点1的位置，每次有p的概率走1步，有1-p的概率走2步，
给出n,p,和n个雷的坐标xi,问不踩到地雷的概率
数据范围 ： 1 <= n <= 10  ,0.25 <= p <= 0.75 ,1 <= xi <= 10^8
分析：
显然有雷的点比没有雷的点多得多，所以计算踩到雷的概率要比计算不踩到雷的概率简单
将路分为n段，（1~x1,x1~x2,x2~x3,...,xn-1~xn）单独计算每段踩到雷的概率,
利用乘法原理求出踩到雷的总概率，不踩到雷的概率 = 1 - 踩到雷的概率
dp[i]表示到达i点的概率
dp[i] = p*dp[i-1]+(1-p)*dp[i-2]
坐标数据太大，直接乘肯定不行，这个时候就需要用到矩阵快速幂
上面dp的状态转移方程其实和斐波那契数列的表达式很像不是吗^_^
用一样的原理构造矩阵就好了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;struct Matrix{double mat[2][2];};Matrix operator * (Matrix a,Matrix b){Matrix c;for (int i = 0;i < 2;++i){for (int j = 0;j < 2;++j){c.mat[i][j] = 0;for (int k = 0;k < 2;++k){c.mat[i][j] += a.mat[i][k]*b.mat[k][j];}}}return c;}Matrix operator ^ (Matrix a,ll k)//矩阵幂 {Matrix c;memset(c.mat,0,sizeof(c.mat));    for(int i=0;i<2;i++)c.mat[i][i]=1;//初始化为单位矩阵     //据说任何矩阵乘以单位矩阵其值不会变    for (;k;k>>=1)    {        if (k&1) c = c*a;        a = a*a;    }    return c; }int x[111]; int main(){int n;double p;while (cin>>n>>p){for (int i = 0;i < n;++i) scanf("%d",x+i);sort(x,x+n);double ans = 1.0;Matrix c;c.mat[0][0] = p,c.mat[0][1] = 1.0-p;c.mat[1][0] = 1.0,c.mat[1][1] = 0.0;Matrix a = c^(x[0]-1);ans *= (1-a.mat[0][0]);for (int i = 1;i < n;++i){if (x[i] == x[i-1]) continue;a = c^(x[i]-x[i-1]-1);ans *= (1.0-a.mat[0][0]);}printf("%.7f\n",ans);}return 0;}

POJ 3071

全概率问题：

当前场次要与j比赛的队伍x是哪个？而x能与j比必然是胜过了对手

        dp[i][j]表示第i次比赛j赢的概率

        dp[i][j] += dp[i-1][j]*dp[i-1][t]*p[j][t]

其中t是第i次比赛可能与j相邻的队伍
第奇数个赢家和前一个赢家比赛
第偶数个赢家和后一个赢家比赛 

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;const int N = 8;double dp[N][1<<N];//dp[i][j]表示第i次比赛j赢的概率/*dp[i][j] += dp[i-1][j]*dp[i-1][t]*p[j][t]其中t是第i次比赛可能与j相邻的队伍第奇数个赢家和前一个赢家比赛第偶数个赢家和后一个赢家比赛 */ int n;double p[1<<N][1<<N];int main(){    while (Sint(n),~n)    {    for (int i = 0;i < (1<<n);++i)    {    for (int j = 0;j < (1<<n);++j)    {    SDB(p[i][j]);}}mem(dp,0);for (int i = 0;i < (1<<n);++i) dp[0][i] = 1;for (int i = 1;i <= n;++i)//进行n场比赛{for (int j = 0;j < (1<<n);++j){int t = j>>(i-1);//j是第t个胜者if (t&1)//奇数与j-1比 {for (int k = t*(1<<(i-1))-1;k >= (t-1)*(1<<(i-1));--k){dp[i][j] += dp[i-1][j]*dp[i-1][k]*p[j][k];}} else //偶数与j+1比 {for (int k = (t+1)*(1<<(i-1));k < (t+2)*(1<<(i-1));++k){dp[i][j] += dp[i-1][j]*dp[i-1][k]*p[j][k];} }}} int mx = 0;for (int i = 0;i < (1<<n);++i){if (dp[n][i]>dp[n][mx]) mx = i;}Pintc(mx+1,'\n');}    return 0;}

CodeForces 148D

Pear和Fish玩游戏游戏：
一个袋子里一开始装着w个白球和b个黑球。

从Pear开始，每次轮流随机抽出一个球。如果抽出的球是白色的，则抽出这个球的人立即获胜。

每当一个球被取出后（然后结算获胜情况后），会有另一个球自动滚出来（不算任何人抽的）。

每个人抽球、和自动滚出来的球都是等概率的。那么Pear获胜率是多少呢？

（以上为原题意抽象成的简单摸球概率问题）

dp[i][j]表示Peal摸球时剩余i个白球和j个黑球的胜率

#define mem(a,x) memset(a,x,sizeof(a))#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define esp 1e-7#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}/*dp[i][j]表示Peal摸球时剩余i个白球和j个黑球的胜率 */const int N = 1111;int w,b;double dp[N][N]; void getdp(double &d,double x,double y,int i,int j){if (j>=2) d += y/(x+y)*(y-1.0)/(x+y-1.0)*   x   /(x+y-2.0)*dp[i-1][j-2];if (j>=3) d += y/(x+y)*(y-1.0)/(x+y-1.0)*(y-2.0)/(x+y-2.0)*dp[i][j-3];}int main(){    while (cin>>w>>b)    {    for (int i = 1;i <= w;++i) dp[i][0] = 1;//全白球必胜     for (int j = 1;j <= b;++j) dp[0][j] = 0;//全黑球必输     for (int i = 1;i <= w;++i)    {    for (int j = 1;j <= b;++j)    {    dp[i][j] = i*1.0/(DB)(i+j);    getdp(dp[i][j],i,j,i,j); }}printf("%.9f\n",dp[w][b]);}    return 0;}

POJ 2151

M个问题，T个队伍，冠军至少解决N题
p[i][j]表示队伍i解决j题的概率
求所有队至少解决1题并且冠军至少解决N题的概率
不用管谁是冠军，只有有人解决了N题，冠军一定满足了

ans = P1(每队至少1题) - P2(每队至少1题但不超过n-1题) 

dp[i][j]表示第i个队解出的题目小于等于j的概率

f[i][j][k]表示第i个队在前j题解出k题的概率

f[i][j][k] = f[i][j-1][k-1]*p[i][j]+p[i][j-1][k]*(1-p[i][j])

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Sint3(x,y,z) scanf("%d %d %d",&x,&y,&z)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define _M_ %#define _C_ ^#define _F_ 5#define _S_ 6#define _A_ ~#define esp 1e-7#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int dx[] = {1,-1,0,0};const int dy[] = {0,0,1,-1};//const int dx[] = {-1,-1,-1,0,0,1,1,1};//const int dy[] = {-1,0,1,-1,1,-1,0,1};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}/*M个问题，T个队伍，冠军至少解决N题p[i][j]表示队伍i解决j题的概率求所有队至少解决1题并且冠军至少解决N题的概率不用管谁是冠军，只有有人解决了N题，冠军一定满足了ans = P1(每队至少1题) - P2(每队至少1题但不超过n-1题) dp[i][j]表示第i个队解出的题目小于等于j的概率f[i][j][k]表示第i个队在前j题解出k题的概率f[i][j][k] = f[i][j-1][k-1]*p[i][j]+p[i][j-1][k]*(1-p[i][j]) */const int N = 1111;DB dp,f[2][111],p[N][N];int main(){    int m,t,n;    while (Sint3(m,t,n)&&(m||t||n))    {    DB P1 = 1.0,P2 = 1.0;    for (int i = 1;i <= t;++i)    {    DB _ = 1.0;    for (int j = 1;j <= m;++j)    {    cin>>p[i][j];    _ = _*(1.0-p[i][j]);}P1 = P1*(1.0-_);}for (int i = 1;i <= t;++i){int _ = 0;mem(f,0);f[0][0] = 1.0;for (int j = 1;j <= m;++j){_=_^1;for (int k = 1;k <= m;++k){f[_][k] = p[i][j]*f[_  _C_ 1][k-1]+(1-p[i][j])*f[_ _C_ 1][k];}f[_][0] = f[_ _C_ 1][0]*(1-p[i][j]);}dp = 0.0;for (int j = 1;j < n;++j) dp += f[_][j];P2 *= dp;}PDB3(P1-P2,'\n');}    return 0;}

POJ 2096

题目的状态有点繁琐

bug的种类，系统是否中毒

当发现一个bug，可能是新的种类，可能造成新的系统中毒

一共是4种情况：

dp[i][j]表示已经发现了i种bug且已经发现j个系统有bug所需要的天数

1.dp[i-1][j-1] //新种类的bug 和新的系统 
2.dp[i-1][j] //新种类的bug 已经有bug的系统
3.dp[i][j-1]//已有种类的bug 新系统
4.dp[i][j] //已有种类的bug 已经有bug的系统 

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Sint3(x,y,z) scanf("%d %d %d",&x,&y,&z)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define _M_ %#define _C_ ^#define _F_ 5#define _S_ 6#define _A_ ~#define esp 1e-7#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int dx[] = {1,-1,0,0};const int dy[] = {0,0,1,-1};//const int dx[] = {-1,-1,-1,0,0,1,1,1};//const int dy[] = {-1,0,1,-1,1,-1,0,1};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}/*n 种 bug  s 个系统1.n种bug全部出现2.s个系统全部有bug dp[i][j]表示已经发现了i种bug且已经发现j个系统有bug所需要的天数1.dp[i-1][j-1] //新种类的bug 和新的系统 2.dp[i-1][j] //新种类的bug 已经有bug的系统3.dp[i][j-1]//已有种类的bug 新系统4.dp[i][j] //已有种类的bug 已经有bug的系统 p1 = (n-i)*(s-j)/(n*s) p2 = (n-i)*j/(n*s)p3 = i*(s-j)/(n*s)p4 = i*j/(n*s)*/const int N = 1111;DB dp[N][N];int main(){    int n,s;    while (cin>>n>>s)    {    mem(dp,0);    for (int i = n;i >= 0;--i)    {    for (int j = s;j >= 0;--j)    {    if (i == n&&j == s) continue;    DB p1 = (n-i)*(s-j)*1.0;    DB p2 = (n-i)*j*1.0;    DB p3 = i*(s-j)*1.0;    DB p4 = 1.0*n*s - i*j*1.0;    dp[i][j] = (p1*dp[i+1][j+1]+p2*dp[i+1][j]+p3*dp[i][j+1] + 1.0*n*s)/p4;//    cout<<dp[i][j]<<endl;}}//cout<<dp[0][0]<<endl;PDB4(dp[0][0],'\n');}    return 0;}

HDU 3853

dp[i][j]表示在[i,j]处到达终点的期望

dp[i][j] = p1*dp[i][j] + p2*dp[i][j+1] + p3*dp[i+1][j] + 2

dp[i][j] = (p2*dp[i][j+1] +  p3*dp[i+1][j] + 2)/(1-p1)

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Sint3(x,y,z) scanf("%d %d %d",&x,&y,&z)#define Sll3(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)#define SDB3(x,y,z) scanf("%lf %lf %lf",&x,&y,&z)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define _M_ %#define _C_ ^#define _F_ 5#define _S_ 6#define _A_ ~#define esp 1e-7#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int dx[] = {1,-1,0,0};const int dy[] = {0,0,1,-1};//const int dx[] = {-1,-1,-1,0,0,1,1,1};//const int dy[] = {-1,0,1,-1,1,-1,0,1};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}/*dp[i][j]表示在[i,j]处到达终点的期望dp[i][j] = p1*dp[i][j] + p2*dp[i][j+1] + p3*dp[i+1][j] + 2dp[i][j] = (p2*dp[i][j+1] +  p3*dp[i+1][j] + 2)/(1-p1)*/const int N = 1111;struct node{DB p1,p2,p3;}p[N][N] ;DB dp[N][N];int main(){    int r,c;    while (Sint2(r,c) == 2)    {    for (int i = 1;i <= r;++i)    {    for (int j = 1;j <= c;++j)    {    node &t = p[i][j];    SDB3(t.p1,t.p2,t.p3);}}mem(dp,0);for (int i = r;i >= 1;--i){for (int j = c;j >= 1;--j){if (i==r&&j==c) continue;node &t = p[i][j];if (fabs(t.p1 - 1.0)<esp) continue;dp[i][j] = t.p2*dp[i][j+1] + t.p3*dp[i+1][j] + 2.0;dp[i][j] /= (1.0 - t.p1);}}PDB3(dp[1][1],'\n');}    return 0;}

HDU 4405

飞行棋游戏，到达坐标大于等于n就胜利

有可以从x直接飞到y的点

掷骰子（1-6），问胜利的期望掷骰子次数

从终点往起点推（因为终点的期望是已知的 -- 0）

如果是不能飞的点i，子状态就是（i+j）(1<=j<=6)

可以飞的点i，子状态就是所有可以飞到i点的状态

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Sint3(x,y,z) scanf("%d %d %d",&x,&y,&z)#define Sll3(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)#define SDB3(x,y,z) scanf("%lf %lf %lf",&x,&y,&z)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define _M_ %#define _C_ ^#define _F_ 5#define _S_ 6#define _A_ ~#define esp 1e-7#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int dx[] = {1,-1,0,0};const int dy[] = {0,0,1,-1};//const int dx[] = {-1,-1,-1,0,0,1,1,1};//const int dy[] = {-1,0,1,-1,1,-1,0,1};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}/*dp[i]表示在i点到达胜利的期望if i > n dp[i] = 0如果i点可以飞：dp[i] = dp[i+(1~6)]/6 + 1如果i点不能飞：dp[i] = 所有i点可以飞到的点的期望和除以这些点的个数 */const int N = 100010;vector<int>u[N];//我在想的是连环飞 ?DB dp[N];int main(){    int n,m;    while (cin>>n>>m)    {if ((!n)&&(!m)) break;    mem(u,0);    mem(dp,0);    for (int i = 1,x,y;i <= m;++i)    {    Sint2(x,y);    u[x].push_back(y);//x可以到达y }for (int i = n-1;i >= 0;--i){if (u[i].size()){DB sum = 0.0;DB t = 0;for (int j = 0;j < u[i].size();++j){int v = u[i][j];//cout<<v<<" ";sum += dp[v];++t;}//cout<<endl;dp[i] = sum/t; }else {DB sum = 0.0, t = 0.0;for (int j = 1;j <= 6;++j){sum += dp[i+j];++t;}dp[i] = sum/t + 1.0;}//cout<<i<<" : "<<dp[i]<<endl;}PDB4(dp[0],'\n'); }    return 0;}

HDU 4336

集齐所有卡片的期望卡片数

首先是可能某个零食里面没有卡片有一个概率p0

其次是可能这个卡片是已经拥有的卡片，是不起作用的

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Sint3(x,y,z) scanf("%d %d %d",&x,&y,&z)#define Sll3(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)#define SDB3(x,y,z) scanf("%lf %lf %lf",&x,&y,&z)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define _M_ %#define _C_ ^#define _F_ 5#define _S_ 6#define _A_ ~#define esp 1e-7#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int dx[] = {1,-1,0,0};const int dy[] = {0,0,1,-1};//const int dx[] = {-1,-1,-1,0,0,1,1,1};//const int dy[] = {-1,0,1,-1,1,-1,0,1};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}DB dp[1<<21],p[22];//dp[st]表示该状态下收集所有的卡片的期望//st中1表示该卡片收集到了,0表示没有收集到 int main(){    int n;    while(cin>>n)    {    double p0 = 1.0;//没有卡片的概率for (int i = 0;i < n;++i){SDB(p[i]);p0 -= p[i];} int V = (1<<n)-1;dp[V] = 0;//所有卡片都有了 期望是0for (int i = V-1;i >= 0;--i){DB pz = p0,ans = 0.0;//pz表示获得已经拥有的卡片，相当于没有获得卡片for (int j = 0;j < n;++j){if (i&(1<<j))//j卡片已经有啦{pz += p[j];} else //获得这张卡片 {ans += p[j]*(dp[i^(1<<j)]+1.0);}dp[i] = (ans+pz)/(1.0-pz); }} PDB4(dp[0],'\n');}    return 0;}

HDU 4652（期望）

掷出连续n次相同或不同的期望

猜到到等比，证明出来的也是等比

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Sint3(x,y,z) scanf("%d %d %d",&x,&y,&z)#define Sll3(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)#define SDB3(x,y,z) scanf("%lf %lf %lf",&x,&y,&z)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define _M_ %#define _C_ ^#define _F_ 5#define _S_ 6#define _A_ ~#define esp 1e-7#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int dx[] = {1,-1,0,0};const int dy[] = {0,0,1,-1};//const int dx[] = {-1,-1,-1,0,0,1,1,1};//const int dy[] = {-1,0,1,-1,1,-1,0,1};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}/*dp[i]表示已经掷出i个相同的面的期望dp[n] = 0 //已有n个就不需要了dp[i] = 1/m * dp[i+1] + (m-1)/m*dp[1] + 1    dp[i+1] = 1/m * dp[i+2] + (m-1)/m * dp[1] + 1    两式相减得dp[i+1] - dp[i] = 1/m * (dp[i+2] - d[i+1])可以发现任意连续的两数之差成等比数列，就可以求出dp[0]。不同的面同理 */int main(){    int T;while (cin>>T) {while (T--){int op,m,n;Sint3(op,m,n);DB ans = 1.0;if (op == 0)//同 {DB p  =  (DB)m,_ = (DB)m;for (int i = 1;i <= n-1;++i){ans += _;_*=p;}}else //不同 {DB _ = 1.0;for (int i = 1;i <= n-1;++i){_*=m*1.0/(m-i);ans += _;}}PDB9(ans,'\n');}}    return 0;}

HDU 4035 (树形DP+期望)

这题esp开得不够小会WA ,利用非常繁复的数学知识去证明

#define mem(a,x) memset(a,x,sizeof(a))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define EX2(x) ((x)*(x))#define EX3(x) ((x)*(x)*(x))#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#include<string>#define Sint(n) scanf("%d",&n)#define Sll(n) scanf("%I64d",&n)#define SDB(n) scanf("%lf",&n)#define Schar(n) scanf("%c",&n)#define Schars(s) scanf("%s",s) #define Sint2(x,y) scanf("%d %d",&x,&y)#define Sll2(x,y) scanf("%I64d %I64d",&x,&y)#define SDB2(n,x) scanf("%lf%lf",&n,&x)#define Schar2(n,x) scanf("%c%c",&n,&x)#define Sint3(x,y,z) scanf("%d %d %d",&x,&y,&z)#define Sll3(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)#define SDB3(x,y,z) scanf("%lf %lf %lf",&x,&y,&z)#define Pint(x) printf("%d",x)#define Pllc(x,c) printf("%I64d%c",x,c)#define Pintc(x,c) printf("%d%c",x,c)#define PDB(x,c) printf("%f%c",x,c)#define PDB1(x,c) printf("%.1f%c",x,c)#define PDB2(x,c) printf("%.2f%c",x,c)#define PDB3(x,c) printf("%.3f%c",x,c)#define PDB4(x,c) printf("%.4f%c",x,c)#define PDB5(x,c) printf("%.5f%c",x,c)#define PDB6(x,c) printf("%.6f%c",x,c)#define PDB7(x,c) printf("%.7f%c",x,c)#define PDB8(x,c) printf("%.8f%c",x,c)#define PDB9(x,c) printf("%.9f%c",x,c)#define PCase() printf("Case %d: ",++kas)#define Pcase() printf("case %d: ",++kas) #define esp 1e-9#define mod#define ls i<<1#define rs i<<1|1#define m(i) ((q[i].l+q[i].r)>>1)using namespace std;typedef long long ll;typedef double DB;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int dx[] = {1,-1,0,0};const int dy[] = {0,0,1,-1};//const int dx[] = {-1,-1,-1,0,0,1,1,1};//const int dy[] = {-1,0,1,-1,1,-1,0,1};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}const int N = 10004;DB k[N],e[N];DB A,B,C;//A B C在叶子的时候是初始化vector<int>s[N];void dfs(int node,int fa){DB a = 0,b = 0,c = 0;for (int i = 0;i < s[node].size();++i){int son = s[node][i];if (son == fa) continue;dfs(son,node);a+=A,b+=B,c+=C;}int t = s[node].size();DB p = (1.0 - k[node] - e[node])/t;A = (k[node]+p*a)/(1.0-p*b);B = p/(1-p*b);C = (1-k[node]-e[node]+p*c)/(1-p*b); }int main(){    int T;cin>>T;int kas = 0;    while (T--)    {    int n;cin>>n; mem(s,0);    for (int i = 1,x,y;i < n;++i)    {    Sint2(x,y);    s[x].push_back(y);    s[y].push_back(x);}for (int i = 1;i <= n;++i){SDB2(k[i],e[i]);k[i]/=100.0;e[i]/=100.0;}dfs(1,-1);PCase();if (fabs(A-1)<esp) puts("impossible");else PDB(C/(1.0-A),'\n');}    return 0;}

总的来说概率DP与数学息息相关，基本算是属于数学的范畴