HDU 4487 Maximum Random Walk（概率DP）

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题目链接：点击打开链接

思路：概率DP，用d[i][j][k]表示第i步，走到j点，走过的最大值为k的概率。然后最后用概率乘以最右边走到的点就是期望，期望相加就是答案。

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;// & 0x7FFFFFFFconst int seed = 131;const ll INF64 = ll(1e18);const int maxn = 200 + 10;int T,n,m,id;double d[2][maxn][maxn];double a, b;int main() {    scanf("%d",&T);    while(T--) {        scanf("%d%d%lf%lf",&id,&n,&a,&b);        int m = n;        printf("%d ",id);        n *= 2;        int u = 0;        memset(d[u], 0, sizeof(d[u]));        d[u][m][m] = 1.0;        for(int i = 0; i < m; i++) {            memset(d[u^1], 0, sizeof(d[u^1]));            for(int j = 1; j< n; j++) {                for(int k = 1; k < n; k++) {                    d[u^1][j-1][k] += d[u][j][k] * a;                    d[u^1][j+1][max(k, j+1)] += d[u][j][k] * b;                    d[u^1][j][k] += d[u][j][k] * (1.0 - a - b);                }            }            u ^= 1;        }        double ans = 0;        for(int i = 0; i <= n; i++) {            for(int j = 0; j <= n; j++) {                ans += d[u][i][j] * (j - m);            }        }        printf("%.4f\n",ans);    }    return 0;}

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