[BZOJ3036] 청개구리의 귀착 확률 DP

링크:

#include <stdio.h>
int main()
{
    puts("       [        by     ]  ");
    puts("  ：blog.csdn.net/vmurder/article/details/46467217");
}

문제 풀이:

어, 토폴로지에서 뒤에서 앞으로 쓸면 돼 Qwq

코드:

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 101000
using namespace std;
struct Eli
{
    int v,l,n;
    bool f;
}e[N<<1];
int head[N],cnt,d[N],D[N];
inline void add(int u,int v,int l)
{
    e[++cnt].v=v;
    e[cnt].l=l;
    e[cnt].n=head[u];
    head[u]=cnt;
}
double f[N];
queue<int>q;
bool vis[N];
int n,m;
void bfs()
{
    int i,u,v;
    q.push(n);
    while(!q.empty())
    {
        vis[u=q.front()]=true,q.pop();
        for(i=head[u];i;i=e[i].n)
            if(!vis[v=e[i].v])q.push(v);
    }
}
int main()
{
    int i,u,v;
    int a,b,c;

    scanf("%d%d",&n,&m);
    for(i=1;i<=m;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        add(b,a,c),d[a]++;
    }
    for(i=1;i<=n;i++)D[i]=d[i]?d[i]:1;
    bfs();
    q.push(n);
    while(!q.empty())
    {
        u=q.front(),q.pop(),f[u]/=D[u];
        for(i=head[u];i;i=e[i].n)
        {
            if(vis[u])f[v=e[i].v]+=f[u]+e[i].l;
            if(--d[v]==0)q.push(v);
        }
    }
    printf("%.2lf
",f[1]);
    return 0;
}