HDU4325——二分——Flowers

http://acm.hdu.edu.cn/showproblem.php?pid=4325

/*
upper_bound 找大于a[i]的最近的下标
lower_bound 找大于等于a[i]的最近的下标
1 2 4 4 4 5 6 ...
l r l r l l l ...
此时q = 4
upper_bound 值为4
lower_bound 值为2
l : (4,5]
r : [4,5)
l - r : [4]
 */
/************************************************
* Author        :Powatr
* Created Time  :2015-8-25 15:03:50
* File Name     :F.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

int l[MAXN], r[MAXN];
int main(){
    int T;
    int n, m;
    int q;
    scanf("%d", &T);
    for(int cas = 1; cas <= T; cas++){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++)
            scanf("%d%d", &l[i], &r[i]);
        sort(l + 1, l + n + 1);
        sort(r + 1, r + n + 1);
        printf("Case #%d:\n", cas);
        for(int i = 1; i <= m; i++){
            scanf("%d", &q);
            int ans = (upper_bound(l + 1, l + n + 1, q) - (l + 1)) - (lower_bound(r + 1, r + n + 1, q) - (r + 1));
            printf("%d\n", ans);
        }
    }
        return 0;
}

　　

转载于:https://www.cnblogs.com/zero-begin/p/4757533.html