[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));

}