ACM-ICPC 2018 沈阳赛区网络预赛 F. Fantastic Graph

最新推荐文章于 2019-08-17 16:40:00 发布

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原文链接：http://www.cnblogs.com/tingtin/p/9610097.html

本文探讨了如何通过选择特定边使二分图中所有顶点的度数落在指定区间内的算法挑战。通过DFS深度优先搜索策略，文章提供了一种解决方案，确保最终图形满足条件，实现了从普通二分图到奇幻图形的转变。

"Oh, There is a bipartite graph.""Make it Fantastic."

X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up several edges from the current graph and try to make the degrees of every point to between the two boundaries. If you pick one edge, the degrees of two end points will both increase by one. Can you help X to check whether it is possible to fix the graph?

Input

There are at most

For each test case,The first line contains three integers $\le N \le 20001≤N≤2000,0 \le M \le 20000≤M≤2000,0 \le K \le 60000≤K≤6000 ). Vertices are numbered from 11 to NN.$

The second line contains two numbers $\le L \le R \le 300)(0≤L≤R≤300). The two fantastic numbers.$

Then $\le U \le N,1 \le V \le M)(1≤U≤N,1≤V≤M). It shows that there is a directed edge from UU-th spot to VV-th spot.$

Note. There may be multiple edges between two vertices.

Output

One line containing a sentence. Begin with the case number. If it is possible to pick some edges to make the graph fantastic, output "Yes" (without quote), else output "No" (without quote).

样例输入复制

样例输出复制

Case 1: Yes
Case 2: No

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

一个二分图，M条边，选用M条边的一些，令最后所有点的度数都在[l,r]内，可以Yes 否则 No

 1 #include <iostream>
 2 #include <vector>
 3 #include <cstdio>
 4 #include <cstring>
 5 using  namespace  std;
 6 #define  P pair<int,int>
 7 #define  ph  push_back
 8 #define  ll long long 
 9 #define  M 4400
10 #define  fi  first
11 #define  se second
12 vector<P>ve[M];
13 int num[M],e[M];
14 int n,m,k,l,r;
15 void dfs(int sta,int maxx)
16 {
17     for(int i=0;i<ve[sta].size();i++){
18         P  p=ve[sta][i];
19         if(!e[p.se]){
20             e[p.se]=1;
21             if(num[p.fi]<maxx&&num[sta]<maxx){//遍历所有的边，在maxx的范围内，尽量用边
22                 num[p.fi]++;
23                 num[sta]++;
24                 dfs(p.fi,maxx);
25             }
26         }
27     }
28 }
29 void  init()
30 {
31     for(int i=1;i<=n+m;i++)
32         {
33             num[i]=0;
34             e[i]=0;
35             ve[i].clear();
36         }
37 }
38 int main()
39 {
40     int cnt=0;
41     while(~scanf("%d%d%d",&n,&m,&k)){
42         init();
43         scanf("%d%d",&l,&r);        
44         int x,y;
45         for(int i=0;i<k;i++)
46         {
47             scanf("%d%d",&x,&y);
48             ve[x].ph({y+n,i});
49             ve[y+n].ph({x,i});
50         }
51         dfs(1,r);
52         int flag=1;
53         for(int i=1;i<=n+m;i++)
54         {
55             if(num[i]<l) {//如果最后还有点的度数不满足条件，就No
56                 flag=0;
57                 break;
58             }
59         }
60         printf("Case %d: ",++cnt);
61         printf("%s\n",flag?"Yes":"No");
62     }
63     return 0;
64 }