好久没写过博客了，把以前的博客补一下。

Necklace

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3603 Accepted Submission(s): 1097

Problem Description

SJX has 2*N magic gems.

Input

Multiple test cases.

For each test case, the first line contains two integers

Output

One line per case, an integer indicates that how many gem will become somber at least.

Sample Input

2 1

1 1

3 4

1 1

1 2

1 3

2 1

Sample Output

Author

HIT

Source

2016 Multi-University Training Contest 1

具体怎么写的忘记了

代码:

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <algorithm>
 8 #include <string>
 9 #include <queue>
10 #include <ctime>
11 #include <vector>
12 using namespace std;
13 typedef long long ll;
14 const int INF=0x3f3f3f3f;
15 const int N=20;
16 int vis[N],a[N];
17 int n,m,ret;
18 int mp[N][N],g[N][N],match[N],used[N];
19 bool dfs(int u){
20     for(int v=1;v<=n;++v){
21         if(!g[u][v]||used[v])continue;
22         used[v]=true;
23         if(match[v]==-1||dfs(match[v])){
24             match[v]=u;
25             return true;
26         }
27     }
28     return false;
29 }
30 void solve(){
31     memset(match,-1,sizeof(match));
32     memset(g,0,sizeof(g));
33     for(int i=2;i<=n;++i){
34         for(int j=1;j<=n;++j){
35             if(mp[a[i]][j]||mp[a[i-1]][j])
36                 continue;
37             g[i][j]=true;//把珠子之间的关系做处理，褪色的两个珠子标记为0，不褪色的珠子为1，利用二分图最大匹配找到最大不消退数。
38         }
39     }
40     for(int i=1;i<=n;++i){
41         if(mp[a[1]][i]||mp[a[n]][i])
42             continue;
43         g[1][i]=true;
44     }
45     for(int i=1;i<=n;++i){
46         if(mp[a[1]][i]||mp[a[n]][i])
47             continue;
48         g[1][i]=true;
49     }
50     int ans=0;
51     for(int i=1;i<=n;++i){
52         memset(used,0,sizeof(used));
53         if(dfs(i))++ans;
54     }
55     ret=min(ret,n-ans);
56 }
57 void get(int x){
58     if(ret==0)return;
59     if(x==n+1){
60        solve();return;
61     }
62     for(int i=1;i<=n;i++){
63         if(vis[i])continue;
64         vis[i]=1;
65         a[x]=i;
66         get(x+1);
67         vis[i]=0;
68     }
69 }
70 int main(){
71     int v,u,i;
72     vis[1]=1;a[1]=1;
73     while(~scanf("%d%d",&n,&m)){
74         if(n==0){
75             printf("0\n");
76             continue;
77         }
78         memset(mp,0,sizeof(mp));
79         for(i=0;i<m;i++){
80             scanf("%d%d",&u,&v);
81             mp[v][u]=1;
82         }
83         ret=INF;
84         get(2);//n个阴珠子的全排列(注意:环形序列的全排列)
85         printf("%d\n",ret);
86     }
87     return 0;
88 }

HDU5727.Necklace-二分图匹配匈牙利

Necklace

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