BZOJ1415 聪聪与可可

传送门

题目大意：

一张无向图上有一只猫和一只老鼠，猫先走，鼠后走。猫每次会向与其相邻的并且距离老鼠最近的点移动（若距离相等去编号较小的），如果移动一步后还没吃到老鼠，还可以再移动一步（算在一个时间内的）。老鼠每次会向相邻的点移动或者不移动，所有选择的概率相同。问猫期望多少时间能够吃到老鼠。

题目分析：

期望dp：设(f[i][j])表示猫在i点，鼠在j点，猫吃到鼠的期望步数。(P[i][j])表示与i相邻的点中距离j最近且编号最小的点(bfs预处理)。(deg[i])表示i的度数。
那么：$$f[i][j] = f[P[P[i][j]][j] / (deg[v] + 1) + sum{f[P[P[i][j]][v_j] / (deg[v] + 1)} + 1$$
方程上面两式分别对应：

1. 猫移动，鼠不移动。
2. 猫移动，鼠移动。
   初始化：$$f[i][i] = 0$$

[f[i][j] = 1(dis[i][j] le 2) ]

code

#include<bits/stdc++.h>
using namespace std;
const int N = 1050, M = 1050,OO = 0x3f3f3f3f;
#define eps 1e-8
int n, m, p, q, vt;
int ecnt, adj[N], nxt[M << 1], go[M << 1], P[N][N], dist[N][N], deg[N];
double f[N][N];
int vst[N];
struct node{
    int now, dis;
    node(){}
    node(int _now, int _dis):now(_now), dis(_dis){}
};

inline void addEdge(int u, int v){
    nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v;
    deg[u]++;
}

inline void Bfs(int i){
    static queue<node> que;
    while(!que.empty()) que.pop();
    vt++;
    que.push(node(i, 0));
    vst[i] = vt;
    while(!que.empty()){
        node u = que.front(); que.pop();
        for(int e = adj[u.now]; e; e = nxt[e]){
            int v = go[e];
            if(vst[v] == vt) continue;
            dist[i][v] = u.dis + 1;
            vst[v] = vt;
            node ret = node(v, u.dis + 1);
            que.push(ret);
        }
    }
}

inline void init(){
    memset(dist, OO, sizeof dist);
    for(int i = 1; i <= n; i++) dist[i][i] = 0, Bfs(i);
    for(int i = 1; i <= n; i++){
        P[i][i] = i;
        for(int j = i + 1; j <= n; j++){
            /*与i相邻的与j最近的点*/
            int mndis = OO, point = 0;
            for(int e = adj[i]; e; e = nxt[e]){
                int v = go[e];
                if(dist[v][j] < mndis){
                    mndis = dist[v][j];
                    point = v;
                }
                else if(dist[v][j] == mndis && v < point) point = v;
            }
            P[i][j] = point;
            /*与j相邻的与i最近的点*/
            mndis = OO, point = 0;
            for(int e = adj[j]; e; e = nxt[e]){
                int v = go[e];
                if(dist[v][i] < mndis){
                    mndis = dist[v][i];
                    point = v;
                }
                else if(dist[v][i] == mndis && v < point) point = v;
            }
            P[j][i] = point;
        }
    }
}

inline double F(int u, int v){
    if(f[u][v] >= 0) return f[u][v];
    f[u][v] = 1.0;
    int to = P[P[u][v]][v];
    double p = 1.0 / (1.0*deg[v] + 1.0);
    f[u][v] += F(to, v) * p;
    for(int e = adj[v]; e; e = nxt[e]){
        int vj = go[e];
        f[u][v] += F(to, vj) * p;
    }
    return f[u][v];
}

int main(){
    freopen("h.in", "r", stdin);
    scanf("%d%d%d%d", &n, &m, &p, &q);
    for(int i = 1; i <= m; i++){
        int x, y;
        scanf("%d%d", &x, &y);
        addEdge(x, y), addEdge(y, x);
    }
    init();
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            f[i][j] = -5555;
    for(int i = 1; i <= n; i++){
        f[i][i] = 0;
        for(int j = i + 1; j <= n; j++){
            if(dist[i][j] <= 2) f[i][j] = f[j][i] = 1.0;
        }
    }
    printf("%.3lf
", F(p, q));
}