[BZOJ4832]抵制克苏恩(概率期望DP)

方法一：倒推，最常规的期望DP。f[i][a][b][c]表示还要再攻击k次，目前三种随从个数分别为a,b,c的期望攻击英雄次数，直接转移即可。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,l,r) for (int i=l; i<=r; i++)
typedef long long ll;
using namespace std;

const int N=60,M=10;
int n,a,b,c,T;
double f[N][M][M][M];

int main(){
    freopen("bzoj4832.in","r",stdin);
    freopen("bzoj4832.out","w",stdout);
    rep(i,1,50) rep(a,0,7) rep(b,0,7-a) rep(c,0,7-a-b){
        int t=(a+b+c<7);
        f[i][a][b][c]+=(f[i-1][a][b][c]+1)/(a+b+c+1);
        if (a) f[i][a][b][c]+=f[i-1][a-1][b][c]*a/(a+b+c+1);
        if (b) f[i][a][b][c]+=f[i-1][a+1][b-1][c+t]*b/(a+b+c+1);
        if (c) f[i][a][b][c]+=f[i-1][a][b+1][c-1+t]*c/(a+b+c+1);
    }
    for (scanf("%d",&T); T--; )
        scanf("%d%d%d%d",&n,&a,&b,&c),printf("%.2lf
",f[n][a][b][c]);
    return 0;
}

方法二：用顺推做期望DP，f[x]=(f[k]+w[k][x])*p[k][x]，其中k是所有能到达x的状态，w[k][x]表示这个转移的代价（攻击随从时为0，攻击英雄时为1），p[k][x]是x由k得到的概率（注意不是k转移到x的概率）。

P(x由k得到)=P(k)*P(k转移到x)/P(x)，同时维护p和f即可。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,l,r) for (int i=l; i<=r; i++)
typedef long long ll;
using namespace std;

const int N=60,M=10;
const double eps=1e-10;
int n,a,b,c,T;
double p[N][M][M][M],f[N][M][M][M];

int main(){
    freopen("bzoj4832.in","r",stdin);
    freopen("bzoj4832.out","w",stdout);
    for (scanf("%d",&T); T--; ){
        memset(p,0,sizeof(p)); memset(f,0,sizeof(f));
        scanf("%d%d%d%d",&n,&a,&b,&c); p[0][a][b][c]=1; double ans=0;
        rep(i,0,n-1) rep(a,0,7) rep(b,0,7-a) rep(c,0,7-a-b){
            int t=(a+b+c<7);
            p[i+1][a-1][b][c]+=p[i][a][b][c]*a/(a+b+c+1);
            p[i+1][a+1][b-1][c+t]+=p[i][a][b][c]*b/(a+b+c+1);
            p[i+1][a][b+1][c-1+t]+=p[i][a][b][c]*c/(a+b+c+1);
            p[i+1][a][b][c]+=p[i][a][b][c]/(a+b+c+1);
        }
        rep(i,0,n-1) rep(a,0,7) rep(b,0,7-a) rep(c,0,7-a-b){
            int t=(a+b+c<7); double x=f[i][a][b][c]*p[i][a][b][c];
            if (p[i+1][a-1][b][c]>eps)
                f[i+1][a-1][b][c]+=x*a/((a+b+c+1)*p[i+1][a-1][b][c]);
            if (p[i+1][a+1][b-1][c+t]>eps)
                f[i+1][a+1][b-1][c+t]+=x*b/((a+b+c+1)*p[i+1][a+1][b-1][c+t]);
            if (p[i+1][a][b+1][c-1+t]>eps)
                f[i+1][a][b+1][c-1+t]+=x*c/((a+b+c+1)*p[i+1][a][b+1][c-1+t]);
            if (p[i+1][a][b][c]>eps)
                f[i+1][a][b][c]+=(f[i][a][b][c]+1)*p[i][a][b][c]/((a+b+c+1)*p[i+1][a][b][c]);
        }
        rep(a,0,7) rep(b,0,7-a) rep(c,0,7-a-b) ans+=f[n][a][b][c]*p[n][a][b][c];
        printf("%.2lf
",ans);
    }
    return 0;
}

方法三：同样用顺推，但这里的f是上面的f*p，转移时要考虑期望的定义

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,l,r) for (int i=l; i<=r; i++)
typedef long long ll;
using namespace std;

const int N=60,M=10;
const double eps=1e-10;
int n,a,b,c,T;
double p[N][M][M][M],f[N][M][M][M];

int main(){
    for (scanf("%d",&T); T--; ){
        memset(p,0,sizeof(p)); memset(f,0,sizeof(f));
        scanf("%d%d%d%d",&n,&a,&b,&c); p[0][a][b][c]=1; double ans=0;
        rep(i,0,n-1) rep(a,0,7) rep(b,0,7-a) rep(c,0,7-a-b){
            int t=(a+b+c<7);
            p[i+1][a-1][b][c]+=p[i][a][b][c]*a/(a+b+c+1);
            p[i+1][a+1][b-1][c+t]+=p[i][a][b][c]*b/(a+b+c+1);
            p[i+1][a][b+1][c-1+t]+=p[i][a][b][c]*c/(a+b+c+1);
            p[i+1][a][b][c]+=p[i][a][b][c]/(a+b+c+1);
        }
        rep(i,0,n-1) rep(a,0,7) rep(b,0,7-a) rep(c,0,7-a-b){
            int t=(a+b+c<7);
            if (p[i+1][a-1][b][c]>eps)
                f[i+1][a-1][b][c]+=f[i][a][b][c]*a/(a+b+c+1);
            if (p[i+1][a+1][b-1][c+t]>eps)
                f[i+1][a+1][b-1][c+t]+=f[i][a][b][c]*b/(a+b+c+1);
            if (p[i+1][a][b+1][c-1+t]>eps)
                f[i+1][a][b+1][c-1+t]+=f[i][a][b][c]*c/(a+b+c+1);
            if (p[i+1][a][b][c]>eps)
                f[i+1][a][b][c]+=(f[i][a][b][c]+p[i][a][b][c])/(a+b+c+1);
        }
        rep(a,0,7) rep(b,0,7-a) rep(c,0,7-a-b) ans+=f[n][a][b][c];
        printf("%.2lf
",ans);
    }
    return 0;
}