洛谷 P2850 [USACO06DEC]虫洞Wormholes 题解【负环】【SPFA】

点击量：166

   题面里坑太多了唔？

目录

题目描述

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

 

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself 🙂 .

 

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

 

John在他的农场中闲逛时发现了许多虫洞。虫洞可以看作一条十分奇特的有向边，并可以使你返回到过去的一个时刻（相对你进入虫洞之前）。John的每个农场有M条小路（无向边)连接着N （从1..N标号）块地，并有W个虫洞。其中1<=N<=500,1<=M<=2500,1<=W<=200。 现在John想借助这些虫洞来回到过去（出发时刻之前），请你告诉他能办到吗。 John将向你提供F(1<=F<=5)个农场的地图。没有小路会耗费你超过10000秒的时间，当然也没有虫洞回帮你回到超过10000秒以前。

 

输入输出格式

输入格式：

Line 1: A single integer, F. F farm descriptions follow.

 

Line 1 of each farm: Three space-separated integers respectively: N, M, and W

 

Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.

 

Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

 

输出格式：

Lines 1..F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).

 

输入输出样例

输入样例#1：

2

3 3 1

1 2 2

1 3 4

2 3 1

3 1 3

3 2 1

1 2 3

2 3 4

3 1 8

输出样例#1：

NO

YES

说明

For farm 1, FJ cannot travel back in time.

 

For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

解法：

   bulubula说了一大堆，其实就是让我们找一个负环。题目中没有规定要从哪个点进去，只是说要回到过去，隐藏了要回到出发点这一信息，那么我们就要枚举出发点，做SPFA判负环。因为一旦有负环，从负环上的一点进去，一定可以回到过去。这里不用什么玄学操作，只用管一个点是否进队n次。（我直接用了所有点进队n×n次…）

   这个题目还是比较基础的。

Code：

#include<cstdio>
#include<cstring>
#include<queue>
using std::deque;
struct node
{
    int n,v;
    node *nxt;
    node (int n,int v)
    {
        this->n=n;
        this->v=v;
        nxt=NULL;
    }
    node()
    {
        nxt=NULL;
    }
};
node head[511],*tail[511];
int dis[511],n;
bool vis[511],used[511];
int cnt=0;
bool spfa(int x)
{
    memset(used,0,sizeof(used));
    memset(dis,0x3f,sizeof(dis));
    used[x]=true;
    dis[x]=0;
    deque<int> q;
    q.push_back(x);
    vis[x]=true;
    cnt=0;
    while(!q.empty())
    {
        int k=q.front();
        q.pop_front();
        used[k]=false;
        node *p=&head[k];
        while(p->nxt!=NULL)
        {
            p=p->nxt;
            if(dis[k]+p->v<dis[p->n])
            {
                dis[p->n]=dis[k]+p->v;
                cnt++;
                if(cnt>=n*n)//判负环
                    return true;
                if(!used[p->n])
                {
                    used[p->n]=true;
                    vis[p->n]=true;
                    if(q.empty()||dis[p->n]<dis[q.front()])
                        q.push_front(p->n);
                    else
                        q.push_back(p->n);
                }
            }
        }
    }
    return dis[1]<0;
}
int main()
{
    int m,w,u,v,h;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(dis,0x3f,sizeof(dis));
        scanf("%d%d%d",&n,&m,&h);
        for(int i=1;i<=n;i++)
            tail[i]=&head[i];
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            tail[u]->nxt=new node(v,w);
            tail[u]=tail[u]->nxt;
            tail[v]->nxt=new node(u,w);
            tail[v]=tail[v]->nxt;
        }
        for(int i=1;i<=h;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            tail[u]->nxt=new node(v,-w);
            tail[u]=tail[u]->nxt;
        }
        memset(vis,0,sizeof(vis));
        int flag=0;
        for(int i=1;i<=n;i++)
            if(!vis[i])
                if(spfa(i))
                    flag=1;
        if(flag)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}