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;
}