[网络流24题] 分配问题 （二分图的最佳匹配）

题目链接

题目很简单，就是要求一个二分图的二分匹配
数据范围不大，可以用KM算法，也可以用费用流

#include <bits/stdc++.h>
const int maxn = 105;
const int inf = 0x3f3f3f3f;
using namespace std;
typedef long long ll;
struct note{
    int u, v, w, cost;
    int next;
} e1[maxn * maxn * 2], e2[maxn * maxn * 2];
int head[maxn << 1], cnt;	//两个图可以共用一个head[]数组，减少代码量
int n, s, t;

void add(int u, int v, int w, int c){
    e1[cnt] = (note){u, v, w, c, head[u]};
    e2[cnt] = (note){u, v, w, -c, head[u]};
    head[u] = cnt++;
    e1[cnt] = (note){v, u, 0, -c, head[v]};
    e2[cnt] = (note){v, u, 0, c, head[v]};
    head[v] = cnt++;
}
bool vis[maxn<<1];
int flow[maxn << 1], cost[maxn << 1], pre[maxn << 1], last[maxn << 1];

bool spfa(note e[]){
    for (int i = s; i<=t; i++){
        cost[i] = flow[i] = inf;
        vis[i] = false;
    }
    cost[s] = 0;
    pre[t] = -1;
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = head[u]; ~i; i=e[i].next){
            int v = e[i].v, w = e[i].w, c = e[i].cost;
            if(w > 0 && cost[v] > cost[u] + c){
                cost[v] = cost[u] + c;
                pre[v] = u, last[v] = i;
                flow[v] = min(flow[u], w);
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}
int mincost;
int MCMF(note e[]){
    while(spfa(e)){
        mincost += flow[t] * cost[t];
        int now = t;
        while(now != s){
            e[last[now]].w -= flow[t];
            e[last[now] ^ 1].w += flow[t];
            now = pre[now];
        }
    }
    return mincost;
}

int main()
{
    scanf("%d", &n);
    memset(head, -1, sizeof(head));
    s = 0, t = 2 * n + 1;
    for (int i = 1; i <= n; i++){
        add(s, i, 1, 0), add(i + n, t, 1, 0);
        for (int j = 1; j <= n; j++){
            int w;
            scanf("%d", &w);
            add(i, j + n, 1, w);
        }
    }
    printf("%d
", MCMF(e1));
    mincost = 0;
    printf("%d
", -MCMF(e2));
    return 0;
}