[2018百度之星] 资格赛1003 整数规划 (最大匹配)

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题目链接：http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=820&pid=1003

希望最大化目标函数，事实上满足 Kuhn–Munkres algorithm 求解最小权匹配过程中的顶标的定义，现在就是要计算最大顶标和，而这正好就是最小权匹配 ，按照$x_i$和$y_i$给定的约束条件连边，同时边权置为负值。然后跑最大匹配，输出结果的相反数。

这题一般的KM算法是过不了的，因为网上流传的KM代码都到不了$O(n^3)$。于是去UOJ上扒了一份KM模版……

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#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 405;
int w[N][N];
int lx[N], ly[N];
int lmatch[N], rmatch[N];
bool lvis[N], rvis[N];
int slack[N];
int pre[N];
int n;

int update(int n) {
    int dt = INF, ru;
    for (int j = 1; j <= n; j++)
        if (!rvis[j] && slack[j] < dt) {
            dt = slack[j];
            ru = j;
        }
    for (int i = 1; i <= n; i++) {
        if (lvis[i]) lx[i] -= dt;
        if (rvis[i])
            ly[i] += dt;
        else
            slack[i] -= dt;
    }
    return ru;
}

void match(int& u) {
    for (; u; swap(u, lmatch[pre[u]])) rmatch[u] = pre[u];
}

void bfs(int u, int n) {
    queue<int> q;
    q.push(u);
    lvis[u] = true;
    while (true) {
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            for (int v = 1; v <= n; ++v) {
                int tmp;
                if (rvis[v] || (tmp = lx[u] + ly[v] - w[u][v]) > slack[v])
                    continue;
                pre[v] = u;
                if (!tmp) {
                    if (!rmatch[v]) return match(v);
                    rvis[v] = lvis[rmatch[v]] = true;
                    q.push(rmatch[v]);
                } else
                    slack[v] = tmp;
            }
        }
        u = update(n);
        if (!rmatch[u]) return match(u);
        rvis[u] = lvis[rmatch[u]] = true;
        q.push(rmatch[u]);
    }
}
LL KM(int n) {
    for (int i = 1; i <= n; i++) {
        lmatch[i] = rmatch[i] = lx[i] = ly[i] = 0;
        for (int j = 1; j <= n; j++) {
            lx[i] = max(lx[i], w[i][j]);
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            slack[j] = INF;
            lvis[j] = rvis[j] = false;
        }
        bfs(i, n);
    }
    LL res = 0;
    for (int i = 1; i <= n; i++) {
        res += lx[i] + ly[i];
    }
    return res;
}

signed main() {
  // freopen("in", "r", stdin);
  int T_T, _ = 0;
  scanf("%d", &T_T);
  while(T_T--) {
    scanf("%d",&n);
    memset(w, 0, sizeof(w));
    for(int i = 1; i <= n; i++) {
      for(int j = 1; j <= n; j++) {
        scanf("%d", &w[i][j]);
        w[i][j] = -w[i][j];
      }
    }
    
    printf("Case #%d: %I64d\n", ++_, -KM(n));
  }
  return 0;
}