洛谷P4316 绿豆蛙的归宿

一眼看去，这不是高斯消元吗？

然后发现数据范围是100000...

然后发现是DAG...直接拓扑序递推即可。

边(x, y，z)的贡献是P(x) * z / out[x]

 #include <cstdio>
 #include <queue>
 const int N = ;

 struct Edge {
     int v, nex, len;
 }edge[N << ]; int top;

 int in[N], topo[N], e[N], n, out[N];
 double p[N];

 inline void add(int x, int y, int z) {
     top++;
     edge[top].v = y;
     edge[top].len = z;
     edge[top].nex = e[x];
     e[x] = top;
     return;
 }

 inline void topo_sort() {
     std::queue<int> Q;
     for(int i = ; i <= n; i++) {
         if(!in[i]) {
             Q.push(i);
         }
     }
     int t = ;
     while(!Q.empty()) {
         int x = Q.front();
         Q.pop();
         topo[++t] = x;
         for(int i = e[x]; i; i = edge[i].nex) {
             int y = edge[i].v;
             in[y]--;
             if(!in[y]) {
                 Q.push(y);
             }
         }
     }
     return;
 }

 int main() {
     int m;
     scanf("%d%d", &n, &m);
     for(int i = , x, y, z; i <= m; i++) {
         scanf("%d%d%d", &x, &y, &z);
         add(x, y, z);
         in[y]++;
         out[x]++;
     }
     topo_sort();
     p[] = 1.0;
     double ans = 0.0;
     for(int a = ; a <= n; a++) {
         int x = topo[a];
         for(int i = e[x]; i; i = edge[i].nex) {
             int y = edge[i].v;
             p[y] += p[x] / out[x];
             ans += p[x] * edge[i].len / out[x];
         }
     }
     printf("%.2lf", ans);
     return ;
 }

AC代码