POJ3259-Wormholes

转载请注明出处：優YoU  http://user.qzone.qq.com/289065406/blog/1299075341

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利用虫洞的时光旅行，很有趣的一道题。涉及到图论的知识，关键是构造图，用Bellman-Ford算法找出负权环

Bellman-Ford算法核心在于松弛，具体可以百度，这里就不重复前人的智慧了O(∩_∩)O哈哈~

需要注意的就是输入说明Input这部分，很多人读不懂这段题意：

正权（双向）边部分：
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.

 

负权(单向)边部分：
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.

 1 //Memory Time 
 2 //308K   204MS 
 3 
 4 #include<iostream>
 5 using namespace std;
 6 
 7 int dis[1001];      //源点到各点权值
 8 const int max_w=10001;      //无穷远
 9 
10 class weight
11 {
12 public:
13     int s;
14     int e;
15     int t;
16 }edge[5200];
17 
18 int N,M,W_h; //N (1≤N≤500)fields 顶点数
19              //M (1≤M≤2500)paths 正权双向边
20              //W_h (1≤W≤200) wormholes 虫洞（回溯），负权单向边
21 int all_e;   //边集（边总数）
22 
23 bool bellman(void)
24 {
25     bool flag;
26 
27     /*relax*/
28 
29     for(int i=0;i<N-1;i++)
30     {
31         flag=false;
32         for(int j=0;j<all_e;j++)
33             if(dis[edge[j].e] > dis[edge[j].s] + edge[j].t)
34             {
35                 dis[edge[j].e] = dis[edge[j].s] + edge[j].t;
36                 flag=true;         //relax对路径有更新
37             }
38         if(!flag)   
39             break;  //只要某一次relax没有更新，说明最短路径已经查找完毕，或者部分点不可达，可以跳出relax
40     }
41 
42     /*Search Negative Circle*/
43 
44     for(int k=0;k<all_e;k++)
45         if( dis[edge[k].e] > dis[edge[k].s] + edge[k].t)
46             return true;
47 
48     return false;
49 }
50 int main(void)
51 {
52     int u,v,w;
53 
54     int F;
55     cin>>F;
56     while(F--)
57     {
58         memset(dis,max_w,sizeof(dis));    //源点到各点的初始值为无穷，即默认不连通
59 
60         cin>>N>>M>>W_h;
61 
62         all_e=0;      //初始化指针
63 
64         /*read in Positive Paths*/
65 
66         for(int i=1;i<=M;i++)
67         {
68             cin>>u>>v>>w;
69             edge[all_e].s=edge[all_e+1].e=u;
70             edge[all_e].e=edge[all_e+1].s=v;
71             edge[all_e++].t=w;
72             edge[all_e++].t=w;               //由于paths的双向性，两个方向权值相等，注意指针的移动
73         }
74 
75         /*read in Negative Wormholds*/
76 
77         for(int j=1;j<=W_h;j++)
78         {
79             cin>>u>>v>>w;
80             edge[all_e].s=u;
81             edge[all_e].e=v;
82             edge[all_e++].t=-w;     //注意权值为负
83         }
84 
85         /*Bellman-Ford Algorithm*/
86 
87         if(bellman())
88             cout<<"YES"<<endl;
89         else
90             cout<<"NO"<<endl;
91     }
92     return 0;
93 }