Poj 3259 Wormholes（spfa判负环）

Wormholes

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 42366 Accepted: 15560

传送门

Description

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.

Input

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.

Output

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

Sample Input

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

Sample Output

NO

YES

Hint

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.

Source

USACO 2006 December Gold

/有道翻译2.0/

虫洞

时间限制: 2000毫秒 内存限制: 65536 k

总提交: 42366 接受: 15560

描述

在探索他的许多农场,农场主约翰发现了一些惊人的虫洞。 虫洞非常特殊,因为它是一种单向路径送你到目的地之前进入虫洞! 每个FJ农场组成的 N (1≤ N ≤500)字段方便编号1 . . N , 米 (1≤ 米 ≤2500)路径,和 W (1≤ W ≤200)虫洞。

FJ穿越时光的狂热粉丝,他想做以下几点:开始在一些领域,通过一些路径和虫洞旅行,回到起始时间他最初的离开。 也许他将能够满足自己:)。

帮助FJ找出是否这是可能的,他将为你提供完整的地图 F (1≤ F ≤5)他的农场。 没有路径将超过10000秒,没有虫洞旅行可以带回FJ时间超过10000秒。

输入

1号线:一个整数, F 。 F 农场的描述。

每个农场的第1行:三个空格分隔的整数分别为: N , 米 , W

行2 . . 米 每个农场的+ 1:三个空格分隔的数字( 年代 , E , T )描述,分别为:之间的双向道路 年代 和 E 这需要 T 秒遍历。 两个字段可能被多个连接路径。

行 米 + 2 . . 米 + W 每个农场的+ 1:三个空格分隔的数字( 年代 , E , T )描述,分别为:一个路径的一种方式 年代 来 E 这也将旅行回来 T 秒。

输出

行1 . . F :对于每个农场,输出“YES”如果FJ能实现他的目标,否则输出“不”(不包括引号)。

样例输入

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

样例输出

NO

YES

提示

对于农场1,FJ不能穿越时间。

对于农场2,FJ可以穿越时间的周期1 - > 2 - > 3 - > 1,到达之前回到他的起始位置1秒。 他可以从任何地方开始循环来完成这项工作。

源

USACO 2006 12月黄金

/*
邻接矩阵+spfa判负环.
正权值建图双向.
负权值建图单向.
然后定理:若图中一点入队超过n次就出现了负环（bfs）.
         若路径中多次出现一点就出现了负环（dfs）.
*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
#define MAXN 25001
using namespace std;
int s[MAXN],head[MAXN],n,m,tot,cut,dis[MAXN],k;
bool b[MAXN];
struct data
{
    int v;
    int next;
    int z;
}e[MAXN];
void add(int u,int v,int z)
{
    e[cut].v=v;
    e[cut].next=head[u];
    e[cut].z=z;
    head[u]=cut++;
}
bool spfa()
{
    queue<int>q;
    q.push(1);dis[1]=0;b[1]=true;s[1]++;
    while(!q.empty())
    {
        int u=q.front();q.pop();b[u]=0;
        for(int i=head[u];i!=-1;i=e[i].next)
        {
            int v=e[i].v;
            if(dis[v]>dis[u]+e[i].z)
            {
                dis[v]=dis[u]+e[i].z;
            if(!b[v])
            {
                b[v]=true;
                s[v]++;
                if(s[v]>=n) return 1;
                q.push(v);
            }
            }
        }
    }
    return 0;
}
void init()
{
    cut=0;
    memset(dis,0x7f,sizeof(dis));
    memset(b,0,sizeof(b));
    memset(head,-1,sizeof(head));
    memset(s,0,sizeof(s));
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        int x,y,z;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
            add(y,x,z);
        }
        for(int i=1;i<=k;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,-z);
        }
        if(spfa())
        {
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}