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POJ-3259 Wormholes（判负环，spfa算法）

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厌伴老儒烹瓠叶，强随举子踏槐花。这篇文章主要讲述POJ-3259 Wormholes（判负环，spfa算法）相关的知识，希望能为你提供帮助。
题干：
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 FJs 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  Tseconds 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.
【POJ-3259 Wormholes（判负环，spfa算法）】题目大意：


有若干个虫洞，给出了若干普通路径和其所用时间以及虫洞的路径和其倒回的时间，现问你能否回到出发之前的时间，注意普通路径是双向的，虫洞是单向的。

解题报告：


由题目所给信息已经可以构建一张完整的图了，然后进一步理解题目的意思其实是这张图是否存在负环，因此使用Bellman_Ford或者spfa即可。

AC代码：（邻接表储存图）（266ms）

#include< cstdio>
#include< algorithm>
#include< iostream>
#include< queue>
#include< cstdio>
#include< algorithm>
#include< cstring>
using namespace std;
int n,cnt;
const int MAX = 505;
const int INF = 0x3f3f3f3f;
int dis[MAX],maze[MAX][MAX],cntt[MAX],head[MAX];
bool vis[MAX];
struct Edge
int to,w,ne;
Edge()//没有此构造函数不能写node t这样
Edge(int to,int w,int ne):to(to),w(w),ne(ne)//可以写node(pos,cost)这样

e[200000 + 5]; //数组别开小了
void add(int u,int v,int w)
e[cnt].to = v;
e[cnt].w = w;
e[cnt].ne = head[u];
head[u] = cnt++;

bool spfa(int s)
dis[s]=0; vis[s]=1; //队列初始化,s为起点
int v;
queue< int > q;
q.push(s);
while (!q.empty())//队列非空
v=q.front(); //取队首元素
q.pop();
vis[v]=0; //释放队首结点，因为这节点可能下次用来松弛其它节点，重新入队
for(int i=head[v]; i!=-1; i=e[i].ne)//对所有顶点

if (dis[e[i].to]> dis[v]+e[i].w)
dis[e[i].to] = dis[v]+e[i].w; //修改最短路
if (vis[e[i].to]==0)//如果扩展结点i不在队列中，入队

cntt[e[i].to]++;
vis[e[i].to]=1;
q.push(e[i].to);
if(cntt[e[i].to] > =n) return true;





return false;


void init()
cnt = 0;
memset(vis,0,sizeof(vis));
memset(maze,0,sizeof(maze));
memset(e,0,sizeof(e));
memset(dis,INF,sizeof(dis));
memset(cntt,0,sizeof(cntt));
memset(head,-1,sizeof(head));

int main()

int t;
int M,W,u,v,w;
cin> > t;
while(t--)
init();
scanf("%d%d%d",& n,& M,& W);
for(int i = 1; i< =M; i++)
scanf("%d%d%d",& u,& v,& w);
add(u,v,w); add(v,u,w);

for(int i = 1; i< =W; i++)
scanf("%d%d%d",& u,& v,& w);
add(u,v,-w);

cntt[1] =1;
if(spfa(1)) printf("YES\\n");
else printf("NO\\n");


return 0 ;

AC代码2：（邻接矩阵）（需要判重边！）（1079ms）
#include< cstdio>
#include< algorithm>
#include< iostream>

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