POJ-2406 Bellman解决无向图+负权边/SPFA
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 backT 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.
题目大意：有F个牧场，每个牧场有N个区域，看成顶点也可以，这些区域又有M条路（无向的），这些路径都有一定的花费时间，有W个虫洞（单向边），这些虫洞可以穿梭返回t秒前，就问有没有存在一条路径使其能够回到提前起点（举个栗子比如2--->3--->4花费的时间为30s，4--->2之间有一个虫洞能够回溯的时间为31s，那就满足题意了，如果回溯时间是30s或者小于30s，那就说明这条路不满足题意）
#include
#include
#include
#include
#include
#include
using namespace std;
#define rep(i,a,b) for(int i=a;i<=b;i++)
const int INF=0x3f3f3f3f;
struct node
{
int u,v,w;
}num[2550];
int dis[600];
int n;
int e[600][600];
int main()
{
int F;
scanf("%d",&F);
while(F--)
{
memset(num,0,sizeof num);
//memset(e,INF,sizeof e);
memset(dis,0,sizeof dis);
int m,t;
scanf("%d%d%d",&n,&m,&t);
rep(i,1,m)
{
int x,y;
scanf("%d%d%d",&num[i].u,&num[i].v,&num[i].w);
}
rep(i,m+1,m+t)
{
int x,y,z;
scanf("%d%d%d",&num[i].u,&num[i].v,&num[i].w);
num[i].w=-num[i].w;//将虫洞看成负边权
}
rep(i,1,n)
dis[i]=INF;dis[1]=0;
rep(k,1,n-1)
{
bool flag=0;
rep(i,1,m)
{
if(dis[num[i].u]>dis[num[i].v]+num[i].w)//因为不确定这个无向边到底是
//从哪个点到哪个点，所以就要将两个方向都看一下
dis[num[i].u]=dis[num[i].v]+num[i].w,flag=true;
if(dis[num[i].v]>dis[num[i].u]+num[i].w)
dis[num[i].v]=dis[num[i].u]+num[i].w,flag=true;
}
rep(j,m+1,m+t)
if(dis[num[j].v]>dis[num[j].u]+num[j].w)
dis[num[j].v]=dis[num[j].u]+num[j].w,flag=1;
if(!flag) break;
}
bool fl=0;
rep(i,1,m+t)
if(dis[num[i].v]>dis[num[i].u]+num[i].w) fl=1;
if(fl)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
SPFA
#include
#include
#include
#include
#define _clr(x,a) memset(x,a,sizeof(x));
using namespace std;
const int N=600;
const int INF=0x3f3f3f3f;
int t;
int n,m,w;
int u,v,wi;
int g[N][N];
int dis[N];
bool vis[N];
int cnt[N];
void spfa(){
memset(vis,0,sizeof vis);
memset(dis,INF,sizeof dis);
memset(cnt,0,sizeof cnt);
vis[1]=true;
dis[1]=0;
cnt[1]=1;
queue q;
q.push(1);
while(!q.empty()){
int cur=q.front();
q.pop();
for(int i=1;i<=n;i++){
if(dis[i]>dis[cur]+g[cur][i]){
dis[i]=dis[cur]+g[cur][i];
if(!vis[i]){
cnt[i]++;
//判断负环 第n+1次加入队列
if(cnt[i]>n){
printf("YES\n");
return ;
}
q.push(i);
vis[i]=true;
}
}
}
vis[cur]=false;
}
printf("NO\n");
}
int main(void){
//freopen("data.txt","r",stdin);
scanf("%d",&t);
while(t--){
_clr(g,INF);
scanf("%d%d%d",&n,&m,&w);
while(m--){
scanf("%d%d%d",&u,&v,&wi);
if(wi