洛谷P4206 聪聪与可可

无向简单图上给定s，t。每秒s先向t按照最短路走两步(优先节点编号较小的)，然后t随机行动一步。

问期望多少秒相遇。n <= 1000

解：

这个s太蛇皮了...所以预处理一波。

然后不会，看题解发现是SB记忆化搜索......

#include <bits/stdc++.h>

const int N = 1010;
const double eps = 1e-13;

struct Edge {
    int nex, v;
}edge[N << 1]; int tp;

int e[N], gt[N][N], n, m, fr[N], d[N], vis[N], in[N], frfr[N];
double f[N][N];
std::queue<int> Q;

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

void DFS(int x, int s) {
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        if(fr[y] == x) {
            gt[s][y] = gt[s][x];
            DFS(y, s);
        }
    }
    return;
}

inline void BFS(int s) {
    Q.push(s);
    vis[s] = s;
    d[s] = 0;
    fr[s] = 0; frfr[s] = 0;
    while(Q.size()) {
        int x = Q.front();
        Q.pop();
        for(int i = e[x]; i; i = edge[i].nex) {
            int y = edge[i].v;
            if(vis[y] != s) {
                vis[y] = s;
                d[y] = d[x] + 1;
                fr[y] = x;
                frfr[y] = (x == s) ? y : frfr[x];
                Q.push(y);
            }
            else if(d[y] == d[x] + 1) {
                if(frfr[x] < frfr[y]) {
                    frfr[y] = frfr[x];
                    fr[y] = x;
                }
            }
        }
    }
    /// get gt
    gt[s][s] = s; f[s][s] = 0;
    for(int i = e[s]; i; i = edge[i].nex) {
        int y = edge[i].v;
        gt[s][y] = y;
        DFS(y, s);
    }
    return;
}

double F(int x, int y) {
    if(f[x][y] >= -eps) {
        return f[x][y];
    }
    if(gt[x][y] == y || gt[gt[x][y]][y] == y) {
        return f[x][y] = 1;
    }
    double ans = 0;

    int z = gt[gt[x][y]][y];
    for(int i = e[y]; i; i = edge[i].nex) {
        int w = edge[i].v;
        ans += F(z, w);
    }
    ans += F(z, y);

    return f[x][y] = ans / (in[y] + 1) + 1;
}

int main() {
    int s, t;
    scanf("%d%d", &n, &m);
    scanf("%d%d", &s, &t);
    for(int i = 1, x, y; i <= m; i++) {
        scanf("%d%d", &x, &y);
        add(x, y); add(y, x);
        in[x]++; in[y]++;
    }

    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            f[i][j] = -1;
        }
    }

    for(int i = 1; i <= n; i++) {
        BFS(i);
    }

    /*printf("-----------
");
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            printf("%d ", gt[i][j]);
        }
        printf("
");
    }
    printf("-----------
");
    */

    printf("%.3f
", F(s, t));
    return 0;
}

感觉不会死循环的原因是每次之后s和t的距离至少减少1，至多减少3。按s和t的距离分层的话每一步都走到了不同的层。