Alice and Bob

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Alice and Bob‘s game never ends. Today, they introduce a new game. In this game, both of them have N different rectangular cards respectively. Alice wants to use his cards to cover Bob‘s. The card A can cover the card B if the height of A is not smaller than
B and the width of A is not smaller than B. As the best programmer, you are asked to compute the maximal number of Bob‘s cards that Alice can cover.

Please pay attention that each card can be used only once and the cards cannot be rotated.

Input

The first line of the input is a number T (T <= 40) which means the number of test cases.

For each case, the first line is a number N which means the number of cards that Alice and Bob have respectively. Each of the following N (N <= 100,000) lines contains two integers h (h <= 1,000,000,000) and w (w <= 1,000,000,000) which means the height and
width of Alice‘s card, then the following N lines means that of Bob‘s.

Output

For each test case, output an answer using one line which contains just one number.

Sample Input

2
2
1 2
3 4
2 3
4 5
3
2 3
5 7
6 8
4 1
2 5
3 4 

Sample Output

1
2

Source

2012 ACM/ICPC Asia Regional Changchun Online

题意：Alice 和 Bob  分别有 n 张牌，每张牌有一个长（h）和一个宽（w），Bob把他的牌依次排一排，Alice用他的牌依次去覆盖，求最多可以 覆盖多少，覆盖的条件是长和宽都要大于等于Bob的牌。

题解：贪心。将A,B的牌先按h从小到大排，再按w从小到大排。要使得覆盖的牌数最大，贪心策略是使A牌中还没用

的最小的牌尽量覆盖B中的牌，且最接近它。把当前小于牌Ai.h的B牌的w全部丢进multiset里面，再二分查找最后一个小于等于Ai.w的数，若存在则ans++，删掉这个数。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <set>
#include <queue>

using namespace std;
const int maxn=1e5+5;
struct T {
    int h;
    int w;
    bool operator < (const T & a) const {
        return h<a.h;
    }
};
void read(T * a,int n) {
    for(int i=0; i<n; i++)
        scanf("%d%d",&a[i].h,&a[i].w);
}
T a[maxn];
T b[maxn];
int ans;
int n;
multiset<int> st;
multiset<int>::iterator p;
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        read(a,n);
        read(b,n);
        sort(a,a+n);
        sort(b,b+n);
        st.clear();
        ans=0;
        int j=0;
        for(int i=0; i<n; i++) {
            while(j<n&&b[j].h<=a[i].h)
                st.insert(b[j++].w);
            if(st.empty()) continue;
            p=st.lower_bound(a[i].w);
            if(p==st.end()||*p>a[i].w&&p!=st.begin()) p--;
            if(*p<=a[i].w) {
                ans++;
                st.erase(p);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

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