【网络流24题】No.18 分配问题 （二分图最佳匹配 费用流|KM）

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<queue>
  7 #include<cmath>
  8 using namespace std;
  9 #define Maxn 1010
 10 #define INF 0xfffffff
 11 
 12 struct node
 13 {
 14     int x,y,f,o,c,next;
 15 }t[Maxn*Maxn],tt[Maxn*Maxn];int len;
 16 int first[Maxn];
 17 
 18 int mymin(int x,int y) {return x<y?x:y;}
 19 int mymax(int x,int y) {return x>y?x:y;}
 20 
 21 void ins(int x,int y,int f,int c)
 22 {
 23     t[++len].x=x;t[len].y=y;t[len].f=f;t[len].c=c;
 24     t[len].next=first[x];first[x]=len;t[len].o=len+1;
 25     t[++len].x=y;t[len].y=x;t[len].f=0;t[len].c=-c;
 26     t[len].next=first[y];first[y]=len;t[len].o=len-1;
 27 }
 28 
 29 int st,ed;
 30 queue<int > q;
 31 int dis[Maxn],pre[Maxn],flow[Maxn];
 32 bool inq[Maxn];
 33 bool bfs()
 34 {
 35     while(!q.empty()) q.pop();
 36     memset(dis,63,sizeof(dis));
 37     memset(inq,0,sizeof(inq));
 38     q.push(st);dis[st]=0;flow[st]=INF;inq[st]=1;
 39     while(!q.empty())
 40     {
 41         int x=q.front();
 42         for(int i=first[x];i;i=t[i].next) if(t[i].f>0)
 43         {
 44             int y=t[i].y;
 45             if(dis[y]>dis[x]+t[i].c)
 46             {
 47                 dis[y]=dis[x]+t[i].c;
 48                 pre[y]=i;
 49                 flow[y]=mymin(flow[x],t[i].f);
 50                 if(!inq[y])
 51                 {
 52                     inq[y]=1;
 53                     q.push(y);
 54                 }
 55             }
 56         }
 57         inq[x]=0;q.pop();
 58     }
 59     if(dis[ed]>=INF-100000) return 0;
 60     return 1;
 61 }
 62 
 63 void output()
 64 {
 65     for(int i=1;i<=len;i+=2)
 66      printf("%d->%d %d %d\n",t[i].x,t[i].y,t[i].f,t[i].c);
 67     printf("\n");
 68 }
 69 
 70 int max_flow()
 71 {
 72     int ans=0,sum=0;
 73     while(bfs())
 74     {
 75         sum+=dis[ed]*flow[ed];
 76         ans+=flow[ed];
 77         int now=ed;
 78         while(now!=st)
 79         {
 80             t[pre[now]].f-=flow[ed];
 81             t[t[pre[now]].o].f+=flow[ed];
 82             now=t[pre[now]].x;
 83         }
 84     }
 85     return sum;
 86 }
 87 
 88 int n;
 89 
 90 void init()
 91 {
 92     scanf("%d",&n);
 93     st=2*n+1;ed=st+1;
 94     for(int i=1;i<=n;i++)
 95      for(int j=1;j<=n;j++)
 96      {
 97          int x;
 98          scanf("%d",&x);
 99          ins(i,j+n,1,x);
100      }
101     for(int i=1;i<=n;i++) ins(st,i,1,0),ins(i+n,ed,1,0);
102 }
103 
104 int main()
105 {
106     init();
107     for(int i=1;i<=len;i++) tt[i]=t[i];
108     int ans;
109     ans=max_flow();
110     printf("%d\n",ans);
111     for(int i=1;i<=len;i++) t[i]=tt[i];
112     for(int i=1;i<=len;i++) t[i].c=-t[i].c;
113     ans=max_flow();
114     printf("%d\n",-ans);
115     return 0;
116 }