[SCOI2010]游戏

[SCOI2010]游戏

思路

很容易看出这是一个二分图匹配问题，对于每一种属性，我们必须找到一个与之可以相应匹配的物品，所以我们建立的两条边，然后从开始匹配，当遇到第一个无法匹配的属性时我们即可输出答案，因为题目要求所有的攻击必选是连续递增的属性。

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

const int N = 1e6 + 10;

int head[N], to[N << 1], nex[N << 1], cnt = 1;

int match[N], n;

bool visit[N];

void add(int x, int y) {
    to[cnt] = y;
    nex[cnt] = head[x];
    head[x] = cnt++;
}

bool find(int rt) {
    for(int i = head[rt]; i; i = nex[i]) {
        if(!visit[to[i]]) {
            visit[to[i]] = 1;
            if(!match[to[i]] || find(match[to[i]])) {
                match[to[i]] = rt;
                return true;
            }
        }
    }
    return false;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    n = read();
    for(int i = 1; i <= n; i++) {
        int x = read(), y = read();
        add(x, i);
        add(y, i);
    }
    for(int i = 1; i < N; i++) {
        memset(visit, 0, sizeof visit);
        if(!find(i)) {
            printf("%d\n", i - 1);
            return 0;
        }
    }
    return 0;
}