POJ 3259 spfa 判断负环（重边）
Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 68627 | Accepted: 25558 |
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
题意是问是否能通过虫洞回到过去；
虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退Ts。
我们把虫洞看成是一条负权路，问题就转化成求一个图中是否存在负权回路；
Code(1)：邻接矩阵，此时需要处理重边的问题
#include
#include
using namespace std;
int n,m,w;
const int N = 505;
const int INF = 0x3f3f3f3f;
int e[N][N];
bool vis[N];
int d[N],num[N];
int q[N*N];
bool spfa(int s){
memset(vis,false,sizeof(vis));
for(int i = 1; i <= n; ++i) d[i] = (i == s ? 0:INF);
vis[s] = true;
int h = 0,t = 0;
q[t++] = s;
while(h < t){
int u = q[h++];
vis[u] = false;
for(int v = 1; v <= n; ++v){
if(e[u][v] < INF){
if(d[v] > d[u] + e[u][v]){
d[v] = d[u] + e[u][v];
if(!vis[v]){
q[t++] = v;
vis[v] = true;
++num[v];
if(num[v] >= n) return true;
}
}
}
}
}
return false;
}
int main()
{
int t,a,b,c;
cin >> t;
while(t--){
memset(num,0,sizeof(num));
cin >> n >> m >> w;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= n; ++j){
if(i!=j) e[i][j] = INF;
else e[i][j] = 0;
}
}
for(int i = 0; i < m; ++i){
cin >> a >> b >> c;
if(e[a][b] > c){
e[a][b] = e[b][a] = c;
}
}
for(int i = 0; i < w; ++i){
cin >> a >> b >> c;
e[a][b] = -c;
}
if(spfa(1))
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}
Code(2)：前向星，不需要考虑重边
#include
#include
using namespace std;
int n,m,w,tol;
const int N = 505;
const int INF = 0x3f3f3f3f;
int head[N];
int d[N];
bool vis[N];
int q[4*N*N];
int num[N];
struct Edge{
int u,v,w,nex;
}e[N*N];
void init(){
tol = 0;
memset(head,-1,sizeof(head));
}
void addEdge(int u,int v, int w){
e[tol].u = u; e[tol].v = v; e[tol].w= w;
e[tol].nex = head[u];
head[u] = tol++;
}
bool spfa(int s){
for(int i = 1; i <= n; ++i) d[i] = (i == s ? 0 : INF),vis[i] = false;
vis[s] = true;
int h = 0, t= 0;
q[t++] = s;
while(h < t){
int u = q[h++];
vis[u] = false;
for(int i = head[u]; i != -1; i = e[i].nex){
int v = e[i].v,w = e[i].w;
if(d[v] > d[u] + w){
d[v] = d[u] + w;
if(!vis[v]){
vis[v] = true;
q[t++] = v;
++num[v];
if(num[v] > n) return true;
}
}
}
}
return false;
}
int main()
{
int t,a,b,c;
cin >> t;
while(t--){
init();
memset(num,0,sizeof(num));
cin >> n >> m >> w;
for(int i = 0; i < m; ++i){
cin >> a >> b >> c;
addEdge(a,b,c);
addEdge(b,a,c);
}
for(int i = 0; i < w; ++i){
cin >> a >> b >> c;
addEdge(a,b,-c);
}
if(spfa(1))
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}