CF1139E Maximize Mex

题目地址：CF1139E Maximize Mex

这其实是一个二分图匹配匈牙利算法的巧妙运用

考虑倒序回答

则由原来的删除改为添加

把 potential 值作为左部，则一共有编号为 \(0~m\) 共 \(m+1\) 个左部点

把 club 作为右部点

对每一个当前存在的对应关系添加一条对应的无向边

每次回答相当于从小到大对每一个左部点进行匹配，直到无法匹配，此时无法匹配的即为当前答案

之后每添加一条边在上次匹配后的图上继续匹配即可

#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 6;
int n, m, d, a[N], b[N], c[N], ans[N], v[N], w[N], f[N];
vector<int> e[N];

bool dfs(int x) {
    for (unsigned int i = 0; i < e[x].size(); i++) {
        int y = e[x][i];
        if (v[y]) continue;
        v[y] = 1;
        if (!~f[y] || dfs(f[y])) {
            f[y] = x;
            return 1;
        }
    }
    return 0;
}

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
    cin >> d;
    for (int i = 1; i <= d; i++) scanf("%d", &c[i]);
    for (int i = 1; i <= d; i++) w[c[i]] = 1;
    for (int i = 1; i <= n; i++)
        if (!w[i]) {
            if (a[i] > m) continue;
            e[a[i]].push_back(b[i] + m);
            e[b[i]+m].push_back(a[i]);
        }
    memset(f, -1, sizeof(f));
    for (int i = d; i; i--) {
        for (int j = ans[i+1]; j <= m; j++) {
            memset(v, 0, sizeof(v));
            if (!dfs(j)) {
                ans[i] = j;
                break;
            }
        }
        if (a[c[i]] > m) continue;
        e[a[c[i]]].push_back(b[c[i]] + m);
        e[b[c[i]]+m].push_back(a[c[i]]);
    }
    for (int i = 1; i <= d; i++) printf("%d\n", ans[i]);
    return 0;
}