hdu3507 Print Article DP+斜率优化

来源：互联网发布：矩阵复制 matlab 编辑：程序博客网时间：2024/05/16 15:12

点击打开链接

题意：就是给一个序列，让你分成很多段，每段按照题目的公式算出答案，要所有的答案加起来最小

思路一：（参考）http://www.cnblogs.com/lidaxin/p/5116577.html

斜率优化。
设f[i]表示前i个数分段后的最小值,则转移式为f[i]=min{f[j]+(sum[i]-sum[j])^2+M}，1<=j<i
进一步得:f[i]=min{ (f[j]+sum[j]^2-2*sum[i]*sum[j])+(sum[i]^2+M) }

设y(j)=f[j]+sum[j]^2，a[i]=sum[i]，x(j)=sum(j)，则f[i]=min{y(j)-2*a[i]*x(j)}+sum[i]^2+M

则要求min p=y+2ax , 单调队列维护下凸包。

while(L<R && q[L+1].y-2*sum[i]*q[L+1].x <= q[L].y-2*sum[i]*q[L].x) L++;now.x = sum[i];now.y = q[L].y-2*sum[i]*q[L].x+2*sum[i]*sum[i]+m;while(L<R && cross(q[R-1],q[R],now)<=0) R--;q[++R] = now;

now.y = y[i] = f[i]+sum[j]^2 = min{ (f[j]+sum[j]^2-2*sum[i]*sum[j])+(sum[i]^2+M) } + sum[i]^2

答案就是 y[n]=q[R].y=f[n]+sum[n]^2 ==> q[R].y-sum[n]^2；

代码一：

#include <bits/stdc++.h>using namespace std;typedef long long ll;#define mem(a) memset(a,0,sizeof(a))#define mp(x,y) make_pair(x,y)const int maxn = 5e5+10;const int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////int n,m,a;int sum[maxn];struct node{int x,y;}now,q[maxn];int cross(node A,node B,node sum){return (B.x-A.x)*(sum.y-A.y) - (sum.x-A.x)*(B.y-A.y);}int main(){while(scanf("%d%d",&n,&m)==2){mem(sum);for(int i=1; i<=n; i++){scanf("%d",&a);sum[i] = sum[i-1]+a;}int L=0,R=0;for(int i=1; i<=n; i++){while(L<R && q[L+1].y-2*sum[i]*q[L+1].x <= q[L].y-2*sum[i]*q[L].x) L++;now.x = sum[i];now.y = q[L].y-2*sum[i]*q[L].x+2*sum[i]*sum[i]+m;while(L<R && cross(q[R-1],q[R],now)<=0) R--;q[++R] = now;}printf("%d\n",q[R].y-sum[n]*sum[n]);}    return 0;}

思路二：参考：http://www.cnblogs.com/kuangbin/archive/2012/08/26/2657650.html

kuangbin大神= = 非常详细

#include <bits/stdc++.h>using namespace std;typedef long long ll;#define mem(a) memset(a,0,sizeof(a))#define mp(x,y) make_pair(x,y)const int maxn = 5e5+10;const int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////int n,m;int sum[maxn],f[maxn],q[maxn];// int slope(int j,int k){ // WA!!! // return ((double)(f[j]+sum[j]*sum[j]-(f[k]+sum[k]*sum[k]))) / ((double)(2*(sum[j]-sum[k])));// }int getUP(int j,int k){    return f[j]+sum[j]*sum[j]-(f[k]+sum[k]*sum[k]);}int getDOWN(int j,int  k){    return 2*(sum[j]-sum[k]);}int main(){//f[i]=min((sum[i]-sum[j])^2+m+f[j])while(scanf("%d%d",&n,&m)==2){sum[0],f[0]=0;q[0]=0;for(int i=1; i<=n; i++){int a; scanf("%d",&a);sum[i] = sum[i-1]+a;}int L=0,R=0;for(int i=1; i<=n; i++){while(L<R && getUP(q[L+1],q[L]) <= sum[i]*getDOWN(q[L+1],q[L])) L++;int j = q[L];f[i] = f[j] + (sum[i]-sum[j])*(sum[i]-sum[j]) + m;while(L<R && (getUP(i,q[R])*getDOWN(q[R],q[R-1]) <= getUP(q[R],q[R-1])*getDOWN(i,q[R]))) R--;q[++R] = i;}// int L=0,R=0;// for(int i=1; i<=n; i++){// while(L<R && slope(q[L+1],q[L]) <= sum[i]) L++;// int j = q[L];// f[i] = f[j] + (sum[i]-sum[j])*(sum[i]-sum[j]) + m;// while(L<R && slope(i,q[R]) <= slope(q[R],q[R-1])) R--;// q[++R] = i;// }printf("%d\n",f[n]);}    return 0;}

0 0