牛客2018多校第五场E-room 最小费用最大流

题意：有n个寝室，每个寝室4个人，现在在搞搬寝室的活动，告诉你每个寝室之前的人员名单，和之后的人员名单，问最少需要几个人要搬寝室。

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。

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------show time----------------*/
int n,m;
const int maxn =;
const int max_edge = ;
struct Edge{
    int to,nxt = -;
    ll val,cost;
};
Edge gEdges[max_edge*];
ll gHead[maxn],gPre[maxn],gDist[maxn],gPath[maxn];
bool in[maxn];
int gCount = ;
void addEdge(int u,int v,ll c,ll w){
    gEdges[gCount].to = v;
    gEdges[gCount].val = c;
    gEdges[gCount].cost = w;
    gEdges[gCount].nxt = gHead[u];
    gHead[u] = gCount++;

    gEdges[gCount].to = u;
    gEdges[gCount].val = ;
    gEdges[gCount].cost = -*w;
    gEdges[gCount].nxt = gHead[v];
    gHead[v] = gCount++;
}

bool spfa(int s,int t){
                    // memset(gDist, inff, sizeof(gDist));

    for(int i=; i<=*n+; i++){
        gDist[i] = inff;
    }
    memset(gPre , - , sizeof(gPre));
    memset(in,false,sizeof(in));
    queue<int>Q;
    Q.push(s);    in[s] = true;
    gDist[s] = ;
    while(!Q.empty()){
        int u = Q.front(); Q.pop();
        in[u] = false;
        for(int e = gHead[u]; e!=-; e = gEdges[e].nxt){
            int v = gEdges[e].to;
            if(gEdges[e].val> && gDist[v] > gDist[u] + gEdges[e].cost){
                gDist[v] = gDist[u] + gEdges[e].cost;
                gPre[v] = u;
                gPath[v] = e;
                if(in[v]==false){
                    Q.push(v);
                    in[v] = true;
                }
            }
        }
    }
    if(gPre[t]==-)return false;
    else return true;
}
ll flow = ;
ll MinCostFlow(int s,int t){
    ll cost = ;
    while(spfa(s,t)){
        ll f = inff;

        for(int u = t; u!=s; u = gPre[u]){
            if(f > gEdges[gPath[u]].val)
                f = gEdges[gPath[u]].val;
        }
        flow += f;
        cost += 1ll*gDist[t] * f;
        for(int u=t; u!=s; u=gPre[u]){
            gEdges[gPath[u]].val -= f;
            gEdges[gPath[u]^].val += f;
        }
    }
    return cost;
}
struct room{
    int a,b,c,d;
}rm[maxn];
int getw(int x,int y){
    int res = ;
    res -= (rm[x].a == rm[y].a ||rm[x].a == rm[y].b||rm[x].a == rm[y].c||rm[x].a == rm[y].d);
    res -= (rm[x].b == rm[y].a ||rm[x].b == rm[y].b||rm[x].b == rm[y].c||rm[x].b == rm[y].d);
    res -= (rm[x].c == rm[y].a ||rm[x].c == rm[y].b||rm[x].c == rm[y].c||rm[x].c == rm[y].d);
    res -= (rm[x].d == rm[y].a ||rm[x].d == rm[y].b||rm[x].d == rm[y].c||rm[x].d == rm[y].d);
    return res;
}
int main(){
    memset(gHead,-,sizeof(gHead));
    int s,t;
    scanf("%d", &n);
    for(int i=; i<=n*; i++){
        scanf("%d%d%d%d", &rm[i].a, &rm[i].b, &rm[i].c, &rm[i].d);
    }

    for(int i=; i<=n; i++){
        addEdge(,i,,);
        addEdge(n+i,*n+,,);
        for(int j=+n; j<=*n; j++){
            addEdge(i,j,,getw(i,j));
        }
    }
    ll cost = MinCostFlow(,*n+);
    printf("%lld\n" ,cost);
    return ;
}

牛客

参考：

作者：东林AotoriChiaki
链接：https://www.nowcoder.com/discuss/90015?type=101&order=0&pos=1&page=0
来源：牛客网