bzoj 1415 期望dp + 记忆化搜索

思路：这个题看着感觉不能dp，其实是可以dp的，因为狼每次走两步，兔子每次走一步，每进行一轮以后，狼和兔子的距离

肯定是在接近的，没有相同的状态，dp之前预处理出来，每一步狼该往哪里走。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N =  + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +;

int n, m, S, T, dis[N][N], nx[N][N];
bool vis[N];
vector<int> edge[N];

double f[N][N];

void bfs(int s) {

    memset(vis, false, sizeof(vis));
    dis[s][s] = ; vis[s] = true;
    queue<int> que;
    for(int i = ; i < edge[s].size(); i++) {
        int v = edge[s][i];
        dis[s][v] = ;
        nx[s][v] = v;
        vis[v] = true;
        que.push(v);
    }

    while(!que.empty()) {
        int u = que.front(); que.pop();
        for(int i = ; i < edge[u].size(); i++) {
            int v = edge[u][i];
            if(!vis[v]) {
                dis[s][v] = dis[s][u] + ;
                nx[s][v] = nx[s][u];
                vis[v] = true;
                que.push(v);
            } else if(vis[v] && dis[s][u] +  == dis[s][v]) {
                nx[s][v] = min(nx[s][v], nx[s][u]);
            }
        }
    }
}

double dp(int i, int j) {
    if(f[i][j] >= ) return f[i][j];
    if(i == j) return f[i][j] = ;
    if(nx[i][j] == j || nx[nx[i][j]][j] == j) return f[i][j] = ;

    f[i][j] = ;
    int nxi = nx[nx[i][j]][j];
    double p = 1.0 / ((int) edge[j].size() + );

    for(int k = ; k < edge[j].size(); k++) {
        f[i][j] += p * dp(nxi, edge[j][k]);
    }

    f[i][j] += p * dp(nxi, j);
    return f[i][j];
}

int main() {

    scanf("%d%d", &n, &m);
    scanf("%d%d", &S, &T);

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

    for(int i = ; i <= m; i++) {
        int u, v; scanf("%d%d", &u, &v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }

    for(int i = ; i <= n; i++) bfs(i);

    printf("%.3f\n", dp(S, T));
    return ;
}

/*
*/