UVA 11383 Golden Tiger Claw 题解

题目

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题解

其实就是一个KM的板子

KM算法在进行中, 需要满足两个点的顶标值之和大于等于两点之间的边权, 所以进行一次KM即可.

KM之后, 顶标之和就是最小的。因为如果不是最小的，就能通过修改顶标值来使某对顶点的顶标值满足\(wx[i]+ly[j]==w[i][j]\)，这样相等子图中又会多一条边，但此时已全部匹配，所以是最小的。

代码

#include <algorithm>
#include <cstdio>
#include <cstring>
#define N 510
using namespace std;
const int inf = 1 << 30;
int s[N][N], lx[N], ly[N], match[N], slack[N], n;
bool vx[N], vy[N];
bool dfs(int u) {
    vx[u] = 1;
    for (int i = 1; i <= n; i++) {
        if (!vy[i]) {
            int t = lx[u] + ly[i] - s[u][i];
            if (t == 0) {
                vy[i] = 1;
                if (match[i] == -1 || dfs(mat[i])) {
                    match[i] = u;
                    return 1;
                }
            } else
                slack[i] = min(slack[i], t);
        }
    }
    return 0;
}
void KM() {
    memset(match, -1, sizeof(match));
    memset(ly, 0, sizeof(ly));
    for (int i = 1; i <= n; i++) {
        lx[i] = -inf;
        for (int j = 1; j <= n; j++) lx[i] = max(lx[i], s[i][j]);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) slack[j] = inf;
        while (1) {
            memset(vx, 0, sizeof(vx));
            memset(vy, 0, sizeof(vy));
            if (dfs(i)) break;
            int d = inf;
            for (int j = 1; j <= n; j++)
                if (!vy[j]) d = min(d, slack[j]);
            for (int j = 1; j <= n; j++)
                if (vx[j]) lx[j] -= d;
            for (int j = 1; j <= n; j++)
                if (vy[j]) ly[j] += d;
        }
    }
    int ans = 0;
    for (int i = 1; i < n; i++) {
        printf("%d ", lx[i]);
        ans += lx[i];
    }
    printf("%d\n", lx[n]);
    ans += lx[n];
    for (int i = 1; i < n; i++) {
        printf("%d ", ly[i]);
        ans += ly[i];
    }
    printf("%d\n", ly[n]);
    ans += ly[n];
    printf("%d\n", ans);
}
int main() {
    while (~scanf("%d", &n)) {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++) scanf("%d", &s[i][j]);
        KM();
    }
    return 0;
}