标签： 二分图最大匹配

2016-10-29 16:23 51人阅读 评论(2) 收藏 举报

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二分图和最大匹配（2） 

Problem Description

There is a kindom of obsession, so people in this kingdom do things very strictly.

They name themselves in integer, and there are n people with their id continuous (s+1,s+2,⋯,s+n) standing in a line in arbitrary order, be more obsessively, people with id x wants to stand at yth position which satisfy

Is there any way to satisfy everyone's requirement?

Input

First line contains an integer T, which indicates the number of test cases.

Every test case contains one line with two integers n, s.

Limits
1≤T≤100.
1≤n≤109.
0≤s≤109.

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the result string.

If there is any way to satisfy everyone's requirement, y equals 'Yes', otherwise y equals 'No'.

Sample Input

2 5 14 4 11

Sample Output

Case #1: No Case #2: Yes

1. #include <iostream>
2. #include <cstdio>
3. #include <algorithm>
4. #include <cmath>
5. #include <queue>
6. #include <vector>
7. #include <cstring>
8. #include <string>
9. using namespace std;
10. const int MAXN=300+7;
11. int n,m;
12. int vis[MAXN],match[MAXN];
13. int num[MAXN];
14. int head[MAXN];
15. int cnt;
16. struct node
17. {
18. int v;
19. int next;
20. } edge[MAXN*MAXN];
21. void add(int u,int v);
22. int dfs(int u);
23. int build()
24. {
25. cnt=0;
26. memset(match,-1,sizeof(match));
27. int i,j,v;
28. for(i=1; i<=n; ++i)head[i]=-1;
29. for(i=1; i<=n; ++i)
30. {
31. match[i]=-1;
32. for(j=1; j<=n; ++j)
33. {
34. v=j+m;
35. if(v%i==0)add(i,j);
36. }
37. }
38. int sum=0;
39. for(i=1; i<=n; ++i)
40. {
41. memset(vis,0,sizeof(vis));
42. if(dfs(i))sum++;
43. }
44. return sum;
45. }
46. void add(int u,int v)
47. {
48. edge[cnt].v=v;
49. edge[cnt].next=head[u];
50. head[u]=cnt++;
51. }
52. int dfs(int u)
53. {
54. int i;
55. for(i=head[u]; i!=-1; i=edge[i].next)
56. {
57. int v=edge[i].v;
58. if(!vis[v])
59. {
60. vis[v]=1;
61. if(match[v]==-1||dfs(match[v]))
62. {
63. match[v]=u;
64. return 1;
65. }
66. }
67. }
68. return 0;
69. }
70. int main()
71. {
72. int t;
73. scanf("%d",&t);
74. for(int t1=1; t1<=t; ++t1)
75. {
76. scanf("%d%d",&n,&m);
77. printf("Case #%d: ",t1);
78. if(m<n)swap(m,n);
79. if(n>300)puts("No");
80. else
81. {
82. int ans=build();
83. if(ans==n)puts("Yes");
84. else puts("No");
85. }
86. }
87. return 0;
88. }