[bzoj1415][Noi2005]聪聪和可可

由于可以求出当老鼠在j点，猫在i点时猫的下一步决策next[i][j]，于是就是水水的期望Dp了。

f[i][j]代表猫在i,老鼠在j时期望几步捉到老鼠。

f[i][j] = (f[next[next[i][j]][j]][j] + ∑(f[next[next[i][j]][j]][v]){(i,v)∈E})/(ind[j] + 1) + 1

View Code

/**************************************************************
    Problem: 1415
    User: lazycal
    Language: C++
    Result: Accepted
    Time:148 ms
    Memory:12800 kb
****************************************************************/
 
#include <cstdio>
#include <queue>
const int N = 1000 + 9;
double f[N][N];
int next[N][N],son[N],n,m,e,c,ind[N],ec;
struct Edge{int link,next;}es[N*2];
inline void addedge(const int x,const int y)
{
    es[++ec].link = y;
    es[ec].next = son[x];
    son[x] = ec;
}
inline void Addedge(const int x,const int y)
{addedge(x,y);addedge(y,x);}
void bfs(const int s)
{
    static std::queue<int> q;
    static int vis[N],time,dis[N];
    ++time; dis[s] = 0;
    for (vis[s] = time,q.push(s); !q.empty(); q.pop()) {
        const int u = q.front();
        for (int i = son[u]; i; i = es[i].next) {
            const int v = es[i].link;
            if (vis[v] == time) {
                if (dis[v] == dis[u] + 1 && next[s][v] > next[s][u]) next[s][v] = next[s][u];
                continue;
            }
            if (u != s) next[s][v] = next[s][u];
            else next[s][v] = v;
            vis[v] = time; dis[v] = dis[u] + 1;
            q.push(v);
        }
    }
}
double dp(const int x,const int y)
{
    if (f[x][y] != -1) return f[x][y];
    if (x == y) return f[x][y] = 0;
    if (next[x][y] == y || next[next[x][y]][y] == y) return f[x][y] = 1;
    f[x][y] = dp(next[next[x][y]][y],y);
    for (int i = son[y]; i; i = es[i].next) f[x][y] += dp(next[next[x][y]][y],es[i].link);
    return (f[x][y] /= ind[y] + 1) += 1;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("1415.in","r",stdin);
    freopen("1415.out","w",stdout);
    #endif
    scanf("%d%d%d%d",&n,&e,&c,&m);
    for (int x,y;e--;) scanf("%d%d",&x,&y),Addedge(x,y),++ind[x],++ind[y];
    for (int i = 1; i <= n; ++i) bfs(i);
    for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) f[i][j] = -1.0;
    printf("%.3f\n",dp(c,m));
}