bzoj 1113

思路：对于两张高度一样的海报 i, j， 即 y[ i ] = y[ j ], 如果对于任意i < k < j 有y[ k ] > y[ i ] && y[ k ] > y[ j ] 那么i 和 j 就能用同一张海报覆盖

那么我们就能用单调栈维护这个过程。 

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

using namespace std;

const int N = 3e5 + 10;
const int M = 10 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int n, x[N], y[N], sk[N], head, rear;
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d%d", &x[i], &y[i]);
    }

    int ans = n;

    for(int i = 1; i <= n; i++) {
        while(head < rear && y[i] <= sk[rear - 1]) {
            if(sk[rear - 1] == y[i]) ans--;
            rear--;
        }

        sk[rear++] = y[i];
    }

    printf("%d
", ans);
    return 0;
}
/*
*/