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 = 100010;

struct Edge {
    int v, nex, len;
}edge[N << 2]; 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 = 1; i <= n; i++) {
        if(!in[i]) {
            Q.push(i);
        }
    }
    int t = 0;
    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 = 1, 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] = 1.0;
    double ans = 0.0;
    for(int a = 1; 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 0;
}