BZOJ2707: [SDOI2012]走迷宫(期望 tarjan 高斯消元)

题意

sol

设\(f[i]\)表示从\(i\)走到\(t\)的期望步数

显然有\(f[x] = \sum_{y} \frac{f[y]}{deg[x]} + 1\)

证明可以用全期望公式。

那么我们可以把每个强联通分量里的点一起高斯消元，就做完了。

(warning：bzoj没有c++11，但是下面的代码是正确的，至于为什么可以点题目链接。。。。)

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, s, t;
int dfn[maxn], low[maxn], vis[maxn], tot, cnt, inder[maxn], col[maxn], ha[maxn];
double f[201][201], deg[maxn], ans[maxn];
stack<int> s;
vector<int> v[maxn], scc[maxn], e[maxn];
void pre() {
    for(int i = 1; i <= n; i++) {
        double sum = 0; f[i][i] = 1.0;
        for(auto &x : v[i]) 
            if(x != t) sum += 1.0 / deg[x], f[i][x] = -1.0;
        f[i][n + 1] = sum;
    }
}
void gauss(int n) {
    for(int i = 1; i <= n; i++) {
        int mx = i;
        for(int j = i + 1; j <= n; j++) if(f[j][i] > f[mx][i]) mx = j;
        if(i != mx) swap(f[i], f[mx]);
        for(int j = 1; j <= n; j++) {
            if(i == j) continue;
            double p = f[j][i] / f[i][i];
            for(int k = i + 1; k <= n + 1; k++) f[j][k] -= f[i][k] * p;
        }
    }
    for(int i = 1; i <= n; i++) f[i][n + 1] = f[i][n + 1] / f[i][i];
}
void tarjan(int x) {
    dfn[x] = low[x] = ++tot; s.push(x); vis[x] = 1;
    for(auto &to : v[x]) {
        if(!dfn[to]) tarjan(to), low[x] = min(low[x], low[to]);
        else if(vis[to]) low[x] = min(low[x], dfn[to]);
    }
    if(low[x] == dfn[x]) {
        int h; cnt++;
        do {
            h = s.top(); s.pop(); 
            vis[h] = 0;
            scc[cnt].push_back(h);
            col[h] = cnt;
        }while(h != x);
    }
}
void solve(vector<int> &p) {
    memset(vis, 0, sizeof(vis));
    memset(f, 0, sizeof(f)); int num = p.size();
    for(int i = 0; i < p.size(); i++) vis[p[i]] = i + 1;
    for(int i = 0; i < p.size(); i++) {
        int x = p[i];
        f[i + 1][i + 1] = deg[x]; f[i + 1][num + 1] = deg[x];
        for(auto &to : v[x]) {
            if(vis[to]) f[i + 1][vis[to]] -= 1;
            else f[i + 1][num + 1] += ans[to];
        }       
    }
    gauss(num);
    for(int i = 0; i < p.size(); i++) ans[p[i]] = f[i + 1][num + 1];
}
void topsort() {
    queue<int> q; q.push(col[t]);
    while(!q.empty()) {
        int p = q.front(); q.pop();
        for(auto &to : e[p]) if(!(--inder[to])) q.push(to);
        if(p != col[t]) 
            solve(scc[p]); 
    }
}
int main() {
    n = read(); m = read(); s = read(); t = read();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read();
        if(x != t) v[x].push_back(y), deg[x]++;
    }
    tarjan(s);
    if(!dfn[t]) {puts("inf"); return 0;}
    for(int i = 1; i <= n; i++) {
        for(auto &x : v[i]) 
            if(col[i] != col[x]) 
                inder[col[i]]++, e[col[x]].push_back(col[i]);
    }
    for(int i = 1; i <= cnt; i++) if(i != col[t] && !inder[i]) {puts("inf"); return 0;}
    topsort();
    printf("%.3lf", ans[s]);
    return 0;
}