牛客OI周赛4-提高组 C 战争(war)

战争(war)

思路:

二分答案, 找到第一个不满足条件的位置

首先对于一个值来说, 所有这个值的区间肯定有交区间, 然后在这个交区间内不能出现比它小的数

所以我们check时从大的值开始考虑, 求出交区间后并标记(用并查集), 如果之后的区间被标记过,那就说明有矛盾

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);

//head

const int N = 5e5 + ;

struct node {

    int l, r, p;

    bool operator < (const node & rhs) & {

        return p > rhs.p;

    }

}a[N], tmp[N];

int fa[N];

int Find(int x) {

    if(x == fa[x]) return x;

    else return fa[x] = Find(fa[x]);

}

int n;

bool check(int m) {

    for (int i = ; i <= m; i++) tmp[i] = a[i];

    for (int i = ; i <= n; i++) fa[i] = i;

    sort(tmp+, tmp++m);

    for (int i = ; i <= m; i++) {

        int j = i, l = tmp[i].l, r = tmp[i].r;

        while(j+ <= m && tmp[j+].p == tmp[i].p) {

            j++;

            l = max(l, tmp[j].l);

            r = min(r, tmp[j].r);

        }

        if(l > Find(r)) return false;

        while(r >= l) {

            if(fa[r] == r) fa[r] = Find(l-), r--;

            else r = Find(r);

        }

        i = j;

    }

    return true;

}

int main() {

    int k;

    scanf("%d %d", &n, &k);

    for (int i = ; i <= k; i++) {

        scanf("%d %d %d", &a[i].l, &a[i].r, &a[i].p);

    }

    int l = , r = k, m = l+r+ >> ;

    while(l < r) {

        if(check(m)) l = m;

        else r = m-;

        m = l+r+ >> ;

    }

    printf("%d\n", m+);

    return ;

}