LightOJ 1364 Expected Cards (概率DP)

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解析：

设dp[c][d][h][s][i][j]为有c,d,h,s个对应的牌，i、j记录大小王的状态。

记忆化搜索一下就ok

[code]:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int C,D,H,S,cnt[5];double dp[15][15][15][15][5][5];void init(){    int c,d,h,s,i,j;    for(c = 0;c < 14;c++)    for(d = 0;d < 14;d++)    for(h = 0;h < 14;h++)    for(s = 0;s < 14;s++)    for(i = 0;i < 5;i++)    for(j = 0;j < 5;j++)        dp[c][d][h][s][i][j] = -1;}double dfs(int c,int d,int h,int s,int i,int j){    if(c > 13||d > 13||h > 13||s > 13) return 0;//printf("status: %d %d %d %d %d %d\n",c,d,h,s,i,j);    cnt[1] = c;    cnt[2] = d;    cnt[3] = h;    cnt[4] = s;    if(i) cnt[i]++;    if(j) cnt[j]++;    if(cnt[1]>=C&&cnt[2]>=D&&cnt[3]>=H&&cnt[4]>=S) return dp[c][d][h][s][i][j] = 0;    if(dp[c][d][h][s][i][j] > -0.5) return dp[c][d][h][s][i][j];    int k,num = cnt[1]+cnt[2]+cnt[3]+cnt[4];    double ans = 0,tmp;    num = 54 - num;    if(num == 0) return 0;    if(!i){        tmp = 100;        for(k = 1;k <= 4;k++){            tmp = min(tmp,dfs(c,d,h,s,k,j));        }        ans += tmp;    }    if(!j){        tmp = 100;        for(k = 1;k <= 4;k++){            tmp = min(tmp,dfs(c,d,h,s,i,k));        }        ans += tmp;    }    ans += (13-c)*dfs(c+1,d,h,s,i,j) + (13-d)*dfs(c,d+1,h,s,i,j)            + (13-h)*dfs(c,d,h+1,s,i,j) + (13-s)*dfs(c,d,h,s+1,i,j);    ans = ans/num+1;    return dp[c][d][h][s][i][j] = ans;}int main(){    int i,j,cas;    scanf("%d",&cas);    for(int T=1;T<=cas;T++){        scanf("%d%d%d%d",&C,&D,&H,&S);        init();        printf("Case %d: ",T);        int x = (C > 13? C - 13 : 0) + (D > 13? D - 13 : 0) + (H > 13? H - 13 : 0) + (S > 13? S - 13 : 0);        if(x > 2) puts("-1");        else{            printf("%.10f\n",dfs(0,0,0,0,0,0));        }    }    return 0;}

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