牛客国庆集训派对Day6 - A 最小费用流

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题目链接:点击这里

解题思路:

(n+1)^2 = n^2 + 2*n + 1, 将根数变成点,一个区域和x根的点的连边花费就是2*x+1，这样跑最小费用流就OK了.

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int mx = 2e2 + 10;
int n,m,K,W,head[mx],tot,S,T;
struct node
{
    int y,f,c;
    int nxt;
}edge[mx*mx];
void AddEdge(int x,int y,int f,int c)
{
    edge[tot] = {y,f,c,head[x]};
    head[x] = tot++;
    edge[tot] = {x,0,-c,head[y]};
    head[y] = tot++;
}
int dis[mx],pre[mx];
bool vis[mx];
bool spfa(){   
    memset(dis,inf,sizeof(dis));   
    memset(pre,-1,sizeof(pre));
    queue<int> que;   que.push(S);   
    dis[S] = 0;    
    while(!que.empty()){       
        int no = que.front();      
        que.pop();     
        vis[no] = 0;       
        for(int i=head[no];~i;i=edge[i].nxt)
        {          
            int y = edge[i].y;         
            if(edge[i].f&&dis[y]>dis[no]+edge[i].c)
            {              
                dis[y] = dis[no] + edge[i].c;              
                pre[y] = i;            
                if(!vis[y])
                {                  
                    que.push(y);                   
                    vis[y] = 1;                
                }          
            }      
        }  
    }  
    return dis[T] != inf;
}
int mincost(){ 
    int ans = 0;   
    while(spfa()){         
        for(int i=pre[T];~i;i=pre[edge[i^1].y])    
        {          
            edge[i].f--;           
            edge[i^1].f++;     
        }      
        ans += dis[T]; 
    }  
    return ans;
}
int main()
{      
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    int a,b,cnt = n+m;
    for(int i=1;i<=n;i++) AddEdge(S,i,1,0);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a,&b);
        AddEdge(i,a+n,1,0);
        AddEdge(i,b+n,1,0);
    }
    for(int i=1;i<=m;i++){
        for(int j=1;j<=n;j++){
            AddEdge(i+n,cnt+j,1,1+(j-1)*2);
        }
    }
    T = n+m+n+1;
    for(int j=1;j<=n;j++){
        AddEdge(cnt+j,T,inf,0);
    }
    printf("%d\n",mincost());
    return 0;
}