POJ2528

Description

思路

板子题,需要离散化。然后从最后一张海报往前枚举看看有没有被完全覆盖。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

typedef long long ll;

const ll maxn = 100010;

ll sum[maxn << 2];
ll lazy[maxn << 2];
vector<int> numb;

void pushup(int rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void pushdown(int rt, int len) {
    if(lazy[rt]) {
        lazy[rt << 1] = lazy[rt];
        lazy[rt << 1 | 1] = lazy[rt];
        sum[rt << 1] = lazy[rt] * (len - (len >> 1));
        sum[rt << 1 | 1] = lazy[rt] * (len >> 1);
        lazy[rt] = 0;
    }
}

void update(int L, int R, int lef, int rig, int rt, ll val) { //将区间[L,R]全设置为1
    if(lef >= L && rig <= R) {
        sum[rt] = val * (rig - lef + 1);
        lazy[rt] = val;
        return ;
    }
    pushdown(rt, rig - lef + 1);
    int mid = (lef + rig) / 2;
    if(L <= mid) update(L, R, lef, mid, rt << 1, val);
    if(R > mid) update(L, R, mid + 1, rig, rt << 1 | 1, val);
    pushup(rt);
}

void build(int lef, int rig, int rt) {
    if(lef > rig) return ;
    lazy[rt] = 0;
    if(lef == rig) {
        sum[rt] = 0;
        return ;
    }
    int mid = (lef + rig) / 2;
    build(lef, mid, rt << 1);
    build(mid + 1, rig, rt << 1 | 1);
    pushup(rt);
}

ll query(int L, int R, int lef, int rig, int rt) {
    if(lef >= L && rig <= R) return sum[rt];
    pushdown(rt, rig - lef + 1);
    int mid = (lef + rig) / 2;
    ll res = 0;
    if(L <= mid) res += query(L, R, lef, mid, rt << 1);
    if(R > mid) res += query(L, R, mid + 1, rig, rt << 1 | 1);
    return res;
}


struct sub {
    int l, r;
}post[20000];

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        numb.clear();
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
            int l, r;
            scanf("%d%d", &l, &r);
            numb.push_back(l);
            numb.push_back(r);
            post[i] = sub{l ,r};
        }
        sort(numb.begin(), numb.end());
        unique(numb.begin(), numb.end());
        for(int i = 0; i < n; i++) {
            post[i].l = lower_bound(numb.begin(), numb.end(), post[i].l) - numb.begin() + 1;
            post[i].r = lower_bound(numb.begin(), numb.end(), post[i].r) - numb.begin() + 1;
        }
        int ans = 0;
        int maxn = numb.size() + 2;
        build(1, maxn, 1);
        for(int i = n - 1; i >= 0; i--) {
            if(query(post[i].l, post[i].r, 1, maxn, 1) < post[i].r - post[i].l + 1) ans++;
            update(post[i].l, post[i].r, 1, maxn, 1, 1);
        }
        printf("%d
", ans);
    }
}
原文地址:https://www.cnblogs.com/limil/p/12698299.html