poj 3259 Wormholes 【spfa判断是否存在负环】
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 34513 | Accepted: 12602 |
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 toF (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 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
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 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：John的农场里有n块地和m条路双向路以及w个虫洞，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T秒。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
说下输入：n块地，m条边，w个虫洞。下面依次是m条边的信息（双向），输入完后是w个虫洞的信息（单向）。
思路：spfa。由于存在负权边，Dijkstra便不能用了。简化题目->就是看图中有没有负权环，有的话John可以无限次走这个环，使得时间一定能得到一个负值。所以存在负环话就是可以，没有的话就是不可以了。
spfa邻接表：157ms
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 500+100
#define MAXM 5000+700
#define INF 0x3f3f3f
using namespace std;
struct Edge
{
int to, val, next;
}edge[MAXM];
int dist[MAXN], vis[MAXM], used[MAXM], head[MAXM], top;
queue<int> Q;
int n, m, w;
void addedge(int a, int b, int d)
{
edge[top].to = b;
edge[top].val = d;
edge[top].next = head[a];
head[a] = top++;
}
void init()
{
top = 0;
for(int i = 1; i <= n; i++)
{
used[i] = 0;
vis[i] = 0;
head[i] = -1;
}
}
void spfa()
{
int i;
for(i = 1; i <= n; i++)
dist[i] = INF;
while(!Q.empty())
{
Q.pop();
}
Q.push(1);
dist[1] = 0;
used[1]++;
vis[1] = 1;
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = 0;//清除标记
for(i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(dist[v] > dist[u] + edge[i].val)
{
dist[v] = dist[u] + edge[i].val;
if(!vis[v])
{
vis[v] = 1;
used[v]++;
if(used[v] > n)//存在负环
{
printf("YES\n");
return ;
}
Q.push(v);
}
}
}
}
printf("NO\n");
}
int main()
{
int t;
int a, b, d;
scanf("%d", &t);
while(t--)
{
scanf("%d%d%d", &n, &m, &w);
init();
while(m--)
{
scanf("%d%d%d", &a, &b, &d);
addedge(a, b, d);
addedge(b, a, d);
}
while(w--)
{
scanf("%d%d%d", &a, &b, &d);
addedge(a, b, -d);//建立负权边
}
spfa();
}
return 0;
}