例题一：POJ-3259 Wormholes(判负环)
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 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 backT 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.
描述
在探索他的许多农场时，约翰农夫发现了许多惊人的虫洞。一个虫洞非常奇特，因为它是一条单向的路径，可以在您进入虫洞之前将它送到目的地！每个FJ的农场包括N（1≤N≤500）方便地编号1..N，M（1≤边号≤2500）路径字段和W（1≤w≤200）虫洞。
由于FJ是一个狂热的时间旅行爱好者，他想要做到以下几点：从某个领域开始，穿过一些路径和虫洞，并在他最初离开之前的一段时间返回起跑场。也许他将能够见到自己:)。
为了帮助FJ发现这是否可行，他会为你提供完整的地图给他的农场的F（1≤F≤5）。没有路径需要超过10,000秒的时间才能移动，并且没有虫洞可以使FJ在时间上超过10,000秒。
题目的大意是给你若干条 正权双向边 和 负权单项边 来建图，问你是否有负权回路。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=100000;
struct node{
int to;
int next;
int w;
}e[maxn];
int flag;
int dis[maxn];
int head[maxn],cn[maxn],sum[maxn],vis[maxn];
int cnt,m,n;
void init()
{
memset(dis,inf,sizeof(dis));
memset(head,-1,sizeof(head));
memset(cn,0,sizeof(cn));
memset(sum,0,sizeof(sum));
memset(vis,0,sizeof(vis));
cnt=0;
}
void add(int x,int y,int d)
{
e[cnt].to=y;
e[cnt].w=d;
e[cnt].next=head[x];
head[x]=cnt++;
}
void spfa()
{
queue<int>q;
q.push(1);
dis[1]=0;
vis[1]=1;
while(!q.empty())
{
int u=q.front();
vis[u]=0;
q.pop();
for(int i=head[u];i!=-1;i=e[i].next)
{
int v=e[i].to;
if(dis[v]>dis[u]+e[i].w)
{
dis[v]=dis[u]+e[i].w;
if(!vis[v])
{
sum[v]++;
vis[v]=1;
q.push(v);
if(sum[v]>=n)
{
flag=1;
break;
}
}
}
}
if(flag)
break;
}
}
int main()
{
int casen,c;
cin>>casen;
while(casen--)
{
init();
flag=0;
scanf("%d%d%d",&n,&m,&c);
int x,y,z;
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&x,&y,&z);
add(x,y,z);
add(y,x,z);
}
for(int i=0;i<c;i++)
{
scanf("%d%d%d",&x,&y,&z);
add(x,y,-z);
}
spfa();
if(flag)
printf("YES
");
else
printf("NO
");
}
return 0;
}
例题二 POJ - 2240
题意：
已知n种货币，以及m种货币汇率及方式，问能否通过货币转换，使得财富增加。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<queue>
5 #include<map>
6 #include<vector>
7 #include<algorithm>
8 #define inf 0x3f3f3f3f
9 using namespace std;
10 typedef long long ll;
11 string s1,s2;
12 const int maxn=1100;
13 int m,n;
14 struct node{
15 int to;
16 int next;
17 double d;
18 }e[maxn];
19 int vis[maxn],f[maxn],head[maxn],cn[maxn];
20 double dis[maxn],w;
21 int cnt,sx;
22 int t=1;
23 int flag;
24 map<string,int>mp;
25 void init()
26 {
27 memset(head,-1,sizeof(head));
28 memset(vis,0,sizeof(vis));
29 memset(dis,0,sizeof(dis));
30 memset(cn,0,sizeof(cn));
31 memset(f,0,sizeof(f));
32 cnt=0;
33 }
34 void add(int x,int y,double w)
35 {
36 e[cnt].to=y;
37 e[cnt].d=w;
38 e[cnt].next=head[x];
39 head[x]=cnt++;
40 }
41 void spfa()
42 {
43 memset(vis,0,sizeof(vis));
44 memset(dis,0,sizeof(dis));
45 memset(cn,0,sizeof(cn));
46 memset(f,0,sizeof(f));
47 flag=0;
48 dis[sx]=1.0;
49 queue<int>q;
50 while(!q.empty())q.pop();
51 q.push(sx);
52 vis[sx]=1;
53 while(!q.empty())
54 {
55 int u=q.front();
56 vis[u]=0;
57 q.pop();
58 if(cn[u]>=n&&vis[u]==0)//只要找到有能成正环的，就一定可以使每个点的值无限增大，一定可以达到指定值，所以为了节省时间，找到正环，即将该点值定义为inf，跳出循环。
59 {
60 dis[u]=inf;
61 vis[u]=1;
62 }
63 if(dis[sx]>1)
64 {
65 flag=1;
66 return;
67 }
68 for(int i=head[u];i!=-1;i=e[i].next)
69 {
70 int v=e[i].to;
71 if(dis[v]<dis[u]*e[i].d)
72 {
73 dis[v]=dis[u]*e[i].d;
74 cn[v]++;
75 if(vis[v]==0)
76 {
77 vis[v]=1;
78 q.push(v);
79 }
80 }
81 }
82 if(flag)
83 break;
84 }
85
86 }
87 int main()
88 {
89
90 while(~scanf("%d",&n))
91 {int id=1;
92 init();
93 flag=0;
94 if(n==0)
95 break;
96 for(int i=1;i<=n;i++)
97 {
98 cin>>s1;
99 mp[s1]=id++;
100 }
101 scanf("%d",&m);
102 for(int i=0;i<m;i++)
103 {
104 cin>>s1>>w>>s2;
105 add(mp[s1],mp[s2],w);
106 }
107 for(int i=1;i<=n;i++)
108 {
109 memset(vis,0,sizeof(vis));
110 memset(dis,0,sizeof(dis));
111 memset(cn,0,sizeof(cn));
112 sx=i;
113 spfa();
114 if(flag)
115 break;
116 }
117 if(flag)
118 {
119 printf("Case %d: Yes
",t++);
120 }
121 else
122 printf("Case %d: No
",t++);
123
124 }
125 return 0;
126 }