[NOI 2005] 聪聪和可可

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=1415

[算法]

        首先BFS预处理出点与点之间的最短路 ， 求出每次聪聪的下一步将会往哪走

        然后 ， 用f[i][j]表示聪聪在i ， 可可在j ， 期望走的步数是多少 ， 概率DP即可

        时间复杂度 : O(N ^ 2)

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010
const int inf = 2e9;

int n , m , s , t , tot;
int deg[MAXN] , head[MAXN];
int nxt[MAXN][MAXN] , dist[MAXN][MAXN];
bool visited[MAXN][MAXN]; 
double f[MAXN][MAXN];

struct edge
{
        int to , nxt;
} e[MAXN << 1];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u , int v)
{
        ++tot;
        e[tot] = (edge){v , head[u]};
        head[u] = tot;
}
inline void bfs(int s)
{
        int l , r;
        static int q[MAXN];
        for (int i = 0; i <= n; i++) dist[s][i] = inf;
        dist[s][s] = 0;
        q[l = r = 1] = s;    
        while (l <= r)
        {
                int cur = q[l++];
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to;
                        if (dist[s][cur] + 1 < dist[s][v])
                        {
                                dist[s][v] = dist[s][cur] + 1;
                                q[++r] = v;        
                        }        
                }    
        }    
}
inline double dp(int s , int t)
{
        if (visited[s][t]) return f[s][t];
        if (s == t) return f[s][t] = 0;
        if (nxt[s][t] == t) return f[s][t] = 1;
        if (nxt[nxt[s][t]][t] == t) return f[s][t] = 1;
        for (int i = head[t]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                f[s][t] += 1.0 / (deg[t] + 1) * (dp(nxt[nxt[s][t]][t] , v) + 1);
        }
        f[s][t] += 1.0 / (deg[t] + 1) * (dp(nxt[nxt[s][t]][t] , t) + 1);
        visited[s][t] = true;
        return f[s][t];
}

int main()
{
        
        read(n); read(m); read(s); read(t);
        for (int i = 1; i <= m; i++)
        {
                int u , v;
                read(u); read(v);
                addedge(u , v);
                addedge(v , u);
                ++deg[u]; ++deg[v];
        }
        for (int i = 1; i <= n; i++) bfs(i);
        for (int i = 1; i <= n; i++)
        {
                for (int j = 1; j <= n; j++)
                {
                        for (int k = head[i]; k; k = e[k].nxt)
                        {
                                int v = e[k].to;
                                if (dist[v][j] < dist[j][nxt[i][j]] || (dist[v][j] == dist[j][nxt[i][j]] && v < nxt[i][j])) 
                                        nxt[i][j] = v;
                        }        
                }        
        }
        printf("%.3lf
" , dp(s , t));
        
        return 0;
    
}