题目链接：
http://acm.hust.edu.cn/vjudge/contest/122685#problem/F
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.
题意：
平面上的N个点，有M条正常的道路，有W条长度为负的道路.
求是否存在负环.
题解：
以下分别用bellman-ford和spfa实现.
代码：
bellman-ford：
#include
#include
#include
#include
#include
#define mid(a,b) ((a+b)>>1)
#define LL long long
#define maxn 6000
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;
int m,n,k;
int edges, u[maxn], v[maxn], w[maxn];
int first[maxn], next[maxn];
int dis[maxn];
void add_edge(int s, int t, int val) {
u[edges] = s; v[edges] = t; w[edges] = val;
next[edges] = first[s];
first[s] = edges++;
}
bool bellman(int s) {
for(int i=1; i<=n; i++) dis[i]=inf; dis[s] = 0;
//for(int i=1; i dis[u[e]]+w[e]){
dis[v[e]] = dis[u[e]] + w[e];
if(i == n) return 0;
}
}
// for(int e=0; e dis[u[e]]+w[e])
// return 0;
return 1;
}
int main(int argc, char const *argv[])
{
//IN;
int t; cin >> t;
while(t--)
{
cin >> n >> m >> k;
edges = 0;
for(int i=1; i<=m; i++) {
int u,v,w; scanf("%d %d %d",&u,&v,&w);
add_edge(u,v,w);
add_edge(v,u,w);
}
for(int i=1; i<=k; i++) {
int u,v,w; scanf("%d %d %d",&u,&v,&w);
add_edge(u,v,-w);
}
bool flag = bellman(1);
if(!flag) puts("YES");
else puts("NO");
}
return 0;
} spfa：
#include
#include
#include
#include
#include
#include
#define mid(a,b) ((a+b)>>1)
#define LL long long
#define maxn 6000
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;
int m,n,k;
int edges, u[maxn], v[maxn], w[maxn];
int first[maxn], next[maxn];
int dis[maxn];
void add_edge(int s, int t, int val) {
u[edges] = s; v[edges] = t; w[edges] = val;
next[edges] = first[s];
first[s] = edges++;
}
queue q;
bool inq[maxn];
int inq_cnt[maxn];
bool spfa(int s) {
memset(inq, 0, sizeof(inq));
memset(inq_cnt, 0, sizeof(inq_cnt));
for(int i=1; i<=n; i++) dis[i] = inf; dis[s] = 0;
while(!q.empty()) q.pop();
q.push(s); inq_cnt[s]++;
while(!q.empty()) {
int p = q.front(); q.pop();
inq[p] = 0;
for(int e=first[p]; e!=-1; e=next[e]) if(dis[v[e]] > dis[p]+w[e]){
dis[v[e]] = dis[p] + w[e];
if(!inq[v[e]]) {
q.push(v[e]);
inq[v[e]] = 1;
inq_cnt[v[e]]++;
if(inq_cnt[v[e]] >= n) return 0;
}
}
}
return 1;
}
int main(int argc, char const *argv[])
{
//IN;
int t; cin >> t;
while(t--)
{
cin >> n >> m >> k;
edges = 0;
memset(first, -1, sizeof(first));
for(int i=1; i<=m; i++) {
int u,v,w; scanf("%d %d %d",&u,&v,&w);
add_edge(u,v,w);
add_edge(v,u,w);
}
for(int i=1; i<=k; i++) {
int u,v,w; scanf("%d %d %d",&u,&v,&w);
add_edge(u,v,-w);
}
bool flag = spfa(1);
if(!flag) puts("YES");
else puts("NO");
}
return 0;
}