Summer Training #10 Div.2 D (最大二分匹配）

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<queue>
 4 using namespace std;
 5 #define captype __int32
 6 const int N = 4005;
 7 
 8 
 9 captype cap[N][N],f[N][N],rest[N];
10 int sNode,eNode,pre[N];
11 
12 
13 void init(){
14     memset(f,0,sizeof(f));
15     memset(cap,0,sizeof(cap));
16 }
17 
18 
19 bool searchPath(int n){//找一条增广路
20     bool vist[N]={0};
21     queue<int>q;
22     int u,v;
23     u=sNode; vist[u]=1;
24     pre[u]=u; rest[u]=1<<30;
25     q.push(u);
26     while(!q.empty()){
27         u=q.front(); q.pop();
28         for(v=1; v<=n; v++)
29         if(!vist[v]&&cap[u][v]-f[u][v]>0)
30         {
31             vist[v]=1; pre[v]=u;
32             if(cap[u][v]-f[u][v]>rest[u])
33               rest[v]=rest[u];
34             else
35                 rest[v]=cap[u][v]-f[u][v];
36             if(v==eNode) return true;
37             q.push(v);
38         }
39     }
40     return false;
41 }
42 captype maxflow(int s,int t,int n){
43     captype ans=0;
44     sNode=s; eNode=t;
45     while(searchPath(n)){
46         ans+=rest[eNode];
47         int v=eNode;
48         while(v!=sNode){
49             int u=pre[v];
50             f[u][v]+=rest[eNode];
51             f[v][u]-=rest[eNode];//给一个回流的机会
52             v=u;
53         }
54     }
55     return ans;
56 }
57 int main(){
58     freopen("C:/Users/fuliujun/Desktop/input.txt","r",stdin);
59     freopen("C:/Users/fuliujun/Desktop/output.txt","w",stdout);
60     int a,b,m,n,a1[100],b1[100],n1,sum=0,day=1;
61     while(scanf("%d",&n)!=EOF){
62         for(int i=0;i<n;i++){
63         scanf("%d%d",&a1[i],&b1[i]);
64         sum+=(b1[i]-a1[i])*2;
65         }
66         n1=34+n;
67         m=sum+32+n;
68         init();
69         for(int i=0;i<32;i++)
70             cap[1][i+2]=1;
71         for(int i=0;i<n;i++)
72             cap[34+i][n1]=1;
73         for(int i=0;i<n;i++){
74             int j=(b1[i]-a1[i])*2;
75             for(int z=1;z<=j;z++)
76             {   int tem=a1[i]*2-15;
77                 cap[tem+z][34+i]=1;
78             }
79         }
80         if(n!=0){
81         printf("On day %d Emma can attend as many as %d parties.\n",day,maxflow(1,n1,n1));
82         }
83         day++;
84     }
85     fclose(stdin);
86     fclose(stdout);
87     return 0;
88 }