POJ 3259 Wormholes【Bellman_ford判断负环】

题意：给出n个点，m条正权的边，w条负权的边，问是否存在负环

因为Bellman_ford最多松弛n-1次， 因为从起点1终点n最多经过n-2个点，即最多松弛n-1次，如果第n次松弛还能成功的话，则说明存在有负环

 1 #include<iostream>  

 2 #include<cstdio>  

 3 #include<cstring> 

 4 #include <cmath> 

 5 #include<stack>

 6 #include<vector>

 7 #include<map> 

 8 #include<set>

 9 #include<queue> 

10 #include<algorithm>  

11 using namespace std;

12 

13 typedef long long LL;

14 const int INF = (1<<30)-1;

15 const int maxn=10005;

16 int d[maxn];

17 int flag;

18 int nodenum,edgenum,ecnt;

19 

20 struct Edge{

21     int u,v,cost;

22 } e[maxn];

23 

24 int Bellman_ford(){

25     for(int i=1;i<=nodenum;i++) d[i]=INF;

26     d[1]=0;

27     

28     for(int i=1;i<nodenum;i++){

29         for(int j=1;j<=ecnt;j++){

30             if(d[e[j].v]>d[e[j].u]+e[j].cost)

31             d[e[j].v]=d[e[j].u]+e[j].cost;

32         }

33     }

34     

35     flag=1;

36     for(int i=1;i<=ecnt;i++){

37         if(d[e[i].v]>d[e[i].u]+e[i].cost){

38             flag=0;

39             break;

40         }

41     }

42     return flag;    

43 }

44 

45 void addedges(int a,int b,int c){

46     e[++ecnt].u=a;

47     e[ecnt].v=b;

48     e[ecnt].cost=c;    

49 }

50 

51 int main(){

52     int m,w,ncase;

53     int a,b,c;

54     scanf("%d",&ncase);

55     while(ncase--){

56         ecnt=0;

57         scanf("%d %d %d",&nodenum,&m,&w);

58         while(m--){

59             scanf("%d %d %d",&a,&b,&c);

60             addedges(a,b,c);

61             addedges(b,a,c);            

62         }

63         while(w--){

64             scanf("%d %d %d",&a,&b,&c);

65             c=-c;

66             addedges(a,b,c);

67         }

68         if(Bellman_ford()) printf("NO\n");

69         else printf("YES\n");    

70     }

71     return 0;

72 }

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