【BZOJ1415】】聪聪和可可 记忆化搜索的概率dp

别的代码我是看不懂了，，渣渣只能看得只能看得懂这份用记忆化搜索的代码，，

感觉真的挺好看懂的。 但是想就有点难想了，但是应该比那种单纯的方程好记些，             但是毕竟我以后还是很难想到啊。。

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<set>#include<map>#include<algorithm>using namespace std;const int maxn = 1e3 + 5, oo = 0x3f3f3f3f;struct Edge{    int to, next;}edge[maxn << 1];int head[maxn];int dist[maxn][maxn];int n, e, s, t;double dp[maxn][maxn];bool vis[maxn][maxn];int tot = 0;void addedge(int u, int v){    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void bfs(int s){    queue<int>q;q.push(s);    dist[s][s] = 0;    while(!q.empty()){        int t = q.front();q.pop();        for(int i = head[t]; i != -1; i = edge[i].next){            int v = edge[i].to;            if(dist[v][s] > dist[t][s] + 1){                dist[v][s] = dist[t][s] + 1;                q.push(v);            }        }    }}double dfs(int a, int b){    if(a == b) return dp[a][b] = 0;    if(vis[a][b]) return dp[a][b];    vis[a][b] = true;    int mins = oo;    int nxt = b;    for(int i = head[b]; i != -1; i = edge[i].next){        int v = edge[i].to;        if(dist[v][a] == mins && v < nxt) nxt = v;         else if(dist[v][a] < mins){            mins = dist[v][a];            nxt = v;        }    }    int nxtt = nxt;    if(nxt == a) return dp[a][b] = 1.0;     for(int i = head[nxt]; i != -1; i = edge[i].next){        int v = edge[i].to;        if(dist[v][a] == mins && v < nxtt) nxtt = v;         else if(dist[v][a] < mins){            mins = dist[v][a];            nxtt = v;        }    }    if(nxtt == a) return dp[a][b] = 1.0;    dp[a][b] = dfs(a, nxtt);    int cnt = 1;    for(int i = head[a]; i != -1; i = edge[i].next){        int v = edge[i].to;        cnt++;        dp[a][b] +=  dfs(v, nxtt);    }    dp[a][b] /= cnt * 1.0;    dp[a][b]++;    return dp[a][b];}int main (void){    scanf("%d%d", &n, &e);    scanf("%d%d", &s, &t);    memset(dist, 0x3f, sizeof(dist));    memset(head, -1, sizeof(head));    int x, y;    for(int i = 0; i < e; i++){        scanf("%d%d", &x, &y);        addedge(x, y);        addedge(y, x);    }    for(int i = 1; i <= n; i++) bfs(i);//这个地方用  了n次bfs来求任意两点之间的最短路    printf("%.3f\n", dfs(t, s));    return 0;}