二分图匹配问题

二分图最大匹配

二分图最大权匹配

#include <bits/stdc++.h>
using namespace std;
const int LEN = 1e3 + 5;
typedef long long ll;
int n, n1, n2, m;
const int oo = 1e9;
ll ans;
namespace KM {
    int qh, qt;
    int w[LEN][LEN], q[LEN];
    int lx[LEN], ly[LEN], pre[LEN], slk[LEN], nxt[LEN], vx[LEN], mat[LEN], vy[LEN];
    void match(int &u) {
        while (u) {
            nxt[u] = pre[u];
            swap(u, mat[pre[u]]);
        }
    }
    void bfs(int s) {
        qh = qt = 0;
        vx[q[++qt] = s] = 1;
        while (1) {
            while (qh < qt) {
                int u = q[++qh];
                for (int v = 1; v <= n; v++) {
                    int tmp;
                    if (vy[v] || (tmp = lx[u] + ly[v] - w[u][v]) > slk[v]) continue;
                    pre[v] = u;
                    if (!tmp) {
                        if (!nxt[v]) return match(v);
                        vy[v] = vx[q[++qt] = nxt[v]] = 1;
                    } else slk[v] = tmp;
                }
            }
            int a = oo;
            for (int i = 1; i <= n; i++) {
                if (!vy[i] && a > slk[i]) a = slk[s = i];           
            }
            for (int i = 1; i <= n; i++) {
                if (vx[i]) lx[i] -= a;
                if (vy[i]) ly[i] += a;
                else slk[i] -= a;
            }
            if (!nxt[s]) return match(s);
            vy[s] = vx[q[++qt] = nxt[s]] = 1;
        }
    }
    void km() {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                slk[j] = oo;
                vx[j] = vy[j] = 0;
                lx[i] = max(lx[i], w[i][j]);
            }
            bfs(i);
        }
        ans = 0;
        for (int i = 1; i <= n; i++) ans += lx[i] + ly[i];
        printf("%lld
", ans);
        for (int i = 1; i < n1; i++) {
            printf("%d ", w[i][mat[i]] ? mat[i] : 0);
        }
        printf("%d
", w[n1][mat[n1]] ? mat[n1] : 0);
    }
}
using namespace KM;
int main() {
    scanf("%d %d %d", &n1, &n2, &m);
    n = max(n1, n2);//左右的点的编号都是从１开始，ｎ为较大的值
    for (int i = 1; i <= m; i++) {
        int x, y, z;//表示的是左边的第ｘ个点和右边的第ｙ个点可以花费ｚ匹配
        scanf("%d %d %d", &x, &y, &z);
        w[x][y] = max(w[x][y], z);
    }
    km();
    return 0;
}