BZOJ 1415 聪聪和可可(期望DP)

我们可以用n次BFS预处理出 to[][]数组，to[i][j]表示聪聪从i点到j点第一步会走哪个点。

那么对于聪聪在i点，可可在j点，聪聪先走，定义dp[i][j]表示步数期望。

那么显然有dp[i][j]=(sigma(dp[p][w])+dp[p][j])/(dee[j]+1)+1.

其中p表示to[to[i][j]][j],w表示j点邻接的点。

边界状态就是 如果i==j，那么dp[i][j]=0. 如果i和j的距离在聪聪的一步之内，那么dp[i][j]=1.

记忆化搜索一下即可。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1005;
//Code begin...

struct Edge{int p, next;}edge[N<<1];
int to[N][N], f[N], dee[N], head[N], cnt=1, n, vis[N];
double dp[N][N];
bool mark[N][N];
queue<int>Q;

void add_edge(int u, int v){
    edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;
}
void bfs(){
    FOR(i,1,n) {
        mem(vis,0); mem(f,0);
        Q.push(i); vis[i]=1;
        while (!Q.empty()) {
            int v=Q.front(); Q.pop();
            for (int j=head[v]; j; j=edge[j].next) {
                int u=edge[j].p;
                if (vis[u]) {
                    if (vis[v]==vis[u]-1) f[u]=min(f[u],f[v]);
                    continue;
                }
                Q.push(u); vis[u]=vis[v]+1;
                if (vis[u]<=2) f[u]=u;
                else f[u]=f[v];
            }
        }
        FOR(j,1,n) to[i][j]=f[j];
    }
}
double dfs(int x, int y){
    if (mark[x][y]) return dp[x][y];
    mark[x][y]=1;
    if (x==y) return dp[x][y]=0;
    if (to[x][y]==y||to[to[x][y]][y]==y) return dp[x][y]=1;
    int p=to[to[x][y]][y];
    double res=dfs(p,y);
    for (int i=head[y]; i; i=edge[i].next) res+=dfs(p,edge[i].p);
    res=res/(dee[y]+1)+1;
    return dp[x][y]=res;
}
int main ()
{
    int m, u, v, s, t;
    scanf("%d%d",&n,&m);
    scanf("%d%d",&s,&t);
    while (m--) scanf("%d%d",&u,&v), add_edge(u,v), add_edge(v,u), ++dee[v], ++dee[u];
    bfs();
    printf("%.3f\n",dfs(s,t));
}