牛客网暑期多校第五场E room(费用流)
题意：有n个宿舍，每个宿舍住4个人，给定去年n个宿舍的人员构成，今年n个4人组合的情况，求最少几个人需要搬寝室。
思路：转化为最小费用最大流解决的二分图问题，对每个去年的宿舍，向每个今年的组合连一条边，权值为1费用为需要搬的人数（4-相同的人数），源点到去年各点，今年各点到汇点，都连一条权值为1费用为0的最大流，跑一次费用流即可。
注意点数和边数，不然会段错误（RE）
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int MAXN = 205;
const int MAXM = 40005;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow,cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
N = n;
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for(int i = 0; i < N; i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
int flow = 0;
cost = 0;
while(spfa(s,t))
{
int Min = INF;
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
{
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}
int n;
struct node{
int a,b,c,d;
}r[205];
int cou(int x,int y)
{
int res=4;
res-=(r[x].a==r[y].a||r[x].a==r[y].b||r[x].a==r[y].c||r[x].a==r[y].d);
res-=(r[x].b==r[y].a||r[x].b==r[y].b||r[x].b==r[y].c||r[x].b==r[y].d);
res-=(r[x].c==r[y].a||r[x].c==r[y].b||r[x].c==r[y].c||r[x].c==r[y].d);
res-=(r[x].d==r[y].a||r[x].d==r[y].b||r[x].d==r[y].c||r[x].d==r[y].d);
return res;
}
int main()
{
scanf("%d",&n);
for (int i=1;i<=2*n;i++)
{
scanf("%d%d%d%d",&r[i].a,&r[i].b,&r[i].c,&r[i].d);
}
init(2*n+2);
for (int i=1;i<=n;i++)
{
addedge(0,i,1,0);
addedge(n+i,2*n+1,1,0);
for (int j=n+1;j<=2*n;j++)
{
addedge(i,j,1,cou(i,j));
}
}
int ans;
minCostMaxflow(0,2*n+1,ans);
printf("%d\n",ans);
return 0;
}
