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| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 5513 | Accepted: 2319 |
Description
Input
Output
Sample Input
2
4
35 M classicism programming
0 M baroque skiing
43 M baroque chess
30 F baroque soccer
8
27 M romance programming
194 F baroque programming
67 M baroque ping-pong
51 M classicism programming
80 M classicism Paintball
35 M baroque ping-pong
39 F romance ping-pong
110 M romance Paintball
Sample Output
3
7
Source
求二分图最大独立集。给字符串都做好映射，然后根据题目中条件，把可能组CP的人之间连边，跑匈牙利算法就行。
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<vector> 5 #include<cstring> 6 #include<string> 7 #include<vector> 8 #include<map> 9 using namespace std; 10 const int mxn=2400; 11 struct stu{ 12 int h;//高度 13 bool se;//性别 14 int m,sp; 15 }a[mxn]; 16 map<string,int> mpm,mpsp; 17 vector<int>e[mxn]; 18 int mcnt=0,spcnt=0; 19 int n; 20 // 21 int link[mxn],vis[mxn]; 22 // 23 void clear(){ 24 mpm.clear(); 25 mpsp.clear(); 26 memset(a,0,sizeof a); 27 memset(link,0,sizeof link); 28 for(int i=0;i<=n;i++) e[i].clear(); 29 mcnt=spcnt=0; 30 } 31 bool dfs(int s){//匈牙利算法 32 int i,j; 33 for(i=0;i<e[s].size();i++){ 34 int v=e[s][i]; 35 if(!vis[v]){ 36 vis[v]=1; 37 if(!link[v] || dfs(link[v])){ 38 link[v]=s; 39 return 1; 40 } 41 } 42 } 43 return 0; 44 } 45 void calc(){ 46 int i,j; 47 int ans=n; 48 for(i=1;i<=n;i++){ 49 if(a[i].se){ 50 memset(vis,0,sizeof vis); 51 if(dfs(i))ans--; 52 } 53 } 54 printf("%d\n",ans); 55 } 56 int main(){ 57 int T; 58 scanf("%d",&T); 59 char sex[5],music[60],sport[60]; 60 while(T--){ 61 scanf("%d",&n); 62 clear(); 63 int i,j; 64 for(i=1;i<=n;i++){ 65 cin>>a[i].h>>sex>>music>>sport; 66 if(sex[0]==‘M‘) a[i].se=1; 67 else a[i].se=0; 68 if(!mpm.count(music)){//映射音乐 69 mcnt++; 70 mpm[music]=mcnt; 71 } 72 a[i].m=mpm[music]; 73 if(!mpsp.count(sport)){//映射运动 74 spcnt++; 75 mpsp[sport]=spcnt; 76 } 77 a[i].sp=mpsp[sport]; 78 // printf("test : %d %d %d\n",a[i].se,a[i].m,a[i].sp); 79 } 80 for(i=1;i<n;i++) 81 for(j=i+1;j<=n;j++){ 82 if((abs(a[i].h-a[j].h)<=40)&&(a[i].se!=a[j].se)&&(a[i].m==a[j].m)&&(a[i].sp!=a[j].sp)){//连边条件 83 e[i].push_back(j); 84 e[j].push_back(i); 85 } 86 } 87 calc(); 88 } 89 return 0; 90 }
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原文地址：http://www.cnblogs.com/SilverNebula/p/5744658.html
