[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=,M=;
 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,,) rep(a,,) rep(b,,-a) rep(c,,-a-b){
         int t=(a+b+c<);
         f[i][a][b][c]+=(f[i-][a][b][c]+)/(a+b+c+);
         if (a) f[i][a][b][c]+=f[i-][a-][b][c]*a/(a+b+c+);
         if (b) f[i][a][b][c]+=f[i-][a+][b-][c+t]*b/(a+b+c+);
         if (c) f[i][a][b][c]+=f[i-][a][b+][c-+t]*c/(a+b+c+);
     }
     for (scanf("%d",&T); T--; )
         scanf("%d%d%d%d",&n,&a,&b,&c),printf("%.2lf\n",f[n][a][b][c]);
     return ;
 }

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

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