The Preliminary Contest for ICPC Asia Nanjing 2019

这题概率dp + 拓扑排序可以写
改天补解释

#include <bits/stdc++.h>
using namespace std;

const int maxn=1e5+10;
vector<int>vec[maxn];
int indeg[maxn],seq[maxn];
double d[maxn],f[maxn];
int N,M,T,tot=0;

void topo()
{
    queue<int>q;
    q.push(1);
    tot=0;
    seq[tot++]=1;
    while (!q.empty()) {
        int u=q.front();
        q.pop();
        for (int i=0;i<vec[u].size();i++) {
            if (--indeg[vec[u][i]]==0) {
                q.push(vec[u][i]);
                seq[tot++]=vec[u][i];
            }
        }
    }
}

int main()
{
    scanf("%d",&T);
    while (T--) {
        scanf("%d%d",&N,&M);
        for (int i=1;i<=N;i++) {
            vec[i].clear();
        }
        memset(indeg,0,sizeof(indeg));
        for (int i=0;i<=N;i++) {
            d[i]=0;
            f[i]=0;
        }
        int x,y;
        for(int i=0;i<M;i++) {
            scanf("%d%d",&x,&y);
            vec[x].push_back(y);
            indeg[y]++;
        }
        topo();
        for (int i=tot-2;i>=0;i--) {
            int u=seq[i];
            int cnt=vec[u].size()+1;
            for (int j=0;j<cnt-1;j++) {
                d[u]+=d[vec[u][j]];
            }    
            d[u]=(d[u]/(double)cnt+1)*(double)cnt/(double)(cnt-1);
        }
        for (int i=tot-2;i>=0;i--) {
            int u=seq[i];
            int cnt=vec[u].size()+1;
            for (int j=0;j<cnt-1;j++) {
                f[u]+=f[vec[u][j]];
            }
            f[u]+=cnt*d[u];
            f[u]=f[u]/(double)(cnt-1);
        }
        printf("%.2lf
",f[1]);
    }
    return 0;
}