POJ 3259 Wormholes（判负圈）

Wormholes

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

题目大意：FJ的农场里出现了一些虫洞，虫洞很奇特，它是单向的并且当FJ进入虫洞时它会将FJ送到他回到之前某个时间点。农场里还有一些道路，这些道路是双向的，通过这些道路会花费一些时间。问FJ能否在农场里遇见以前的自己。

解题思路：将普通道路建双向边，权值为正；将虫洞建单向边，权值为负。然后判断图上有无负圈即可。如果有负圈，说明时间会不断减少，肯定可以遇到以前的自己。

代码如下：

Bellman_ford判负圈

<pre name="code" class="cpp">#include <cstdio>#include <climits>#include <cstring>#define INF INT_MAX / 10#define MOD 10000const int maxn = 505;const int maxm = 5555;int d[maxn];int N,M,W,cnt;struct edge{int from,to,cost;};edge edges[maxm];void add_edge(int from,int to,int cost){edges[cnt].from = from;edges[cnt].to = to;edges[cnt++].cost = cost;}bool find_negative_loop(){memset(d,0,sizeof(d));for(int i = 0;i < N;i++){for(int j = 0;j < cnt;j++){edge e = edges[j];if(d[e.to] > d[e.from] + e.cost){d[e.to] = d[e.from] + e.cost;if(i == N - 1) return true;}}}return false;}int main(){int a,b,c,t;scanf("%d",&t);while(t--){scanf("%d %d %d",&N,&M,&W);cnt = 0;for(int i = 0;i < M;i++){scanf("%d %d %d",&a,&b,&c);add_edge(a,b,c);add_edge(b,a,c);}for(int i = 0;i < W;i++){scanf("%d %d %d",&a,&b,&c);add_edge(a,b,-c);}printf("%s\n",find_negative_loop() ? "YES" : "NO");}return 0;}

SPFA判负圈

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <climits>#define INF INT_MAX#define MOD 10000const int maxn = 505;const int maxm = 5555;typedef std::pair<int,int> P;int first[maxn],d[maxn],inque[maxn],cnt[maxn],e;int next[maxm],v[maxm],w[maxm];int N,M,W;void init(){memset(first,-1,sizeof(first));e = 0;}void add_edge(int from,int to,int cost){v[e] = to;next[e] = first[from];w[e] = cost;first[from] = e++;}std::queue<int> que;bool find_negative_loop(){while(que.size()) que.pop();memset(inque,0,sizeof(inque));memset(cnt,0,sizeof(cnt));for(int i = 0;i < N;i++)que.push(i),d[i] = 0;inque[0] = 1;while(que.size()){int u = que.front();que.pop();inque[u] = 0;for(int i = first[u];i != -1;i = next[i]){if(d[v[i]] > d[u] + w[i]){d[v[i]] = d[u] + w[i];if(!inque[v[i]]){que.push(v[i]),inque[v[i]] = 1;if(++cnt[v[i]] > N) return true;}}}}return false;}int main(){int a,b,c,t;scanf("%d",&t);while(t--){init();scanf("%d %d %d",&N,&M,&W);for(int i = 0;i < M;i++){scanf("%d %d %d",&a,&b,&c);add_edge(--a,--b,c);add_edge(b,a,c);}for(int i = 0;i < W;i++){scanf("%d %d %d",&a,&b,&c);add_edge(a-1,b-1,-c);}printf("%s\n",find_negative_loop() ? "YES" : "NO");}return 0;}