| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11772 | Accepted: 4143 |
Description
To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPFi ≤ 1,000; minSPFi ≤ maxSPFi ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn't tan at all........
The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri cows with lotion. A cow may lotion from only one bottle.
What is the maximum number of cows that can protect themselves while tanning given the available lotions?
Input
* Line 1: Two space-separated integers: C and L
* Lines 2..C+1: Line i describes cow i's lotion requires with two integers: minSPFi and maxSPFi
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri
Output
A single line with an integer that is the maximum number of cows that can be protected while tanning
Sample Input
3 2
3 10
2 5
1 5
6 2
4 1
Sample Output
2
Source
题意：
有c头牛，每头牛有一个区间。有l种防晒霜，每种有一个spf值和对应的瓶数。牛只能用在他区间内的防晒霜。问最多能有多少牛被涂。
思路：
按minspf排序，每次取minspf最高的一头牛，给他安排可以被安排的spf值最高的防晒霜。
因为我们这样排了序之后，能给minspf高的牛用的防晒霜肯定也能给minspf低一些的牛用。
这时候就要尽量用spf值高的。
#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
//#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f int c, l;
const int maxn = ;
struct lotion{
int spf;
int cover;
}lo[maxn];
struct mow{
int minspf, maxspf;
}cow[maxn]; bool cmpcow(mow a, mow b)
{
if(a.minspf == b.minspf){
return a.maxspf < b.maxspf;
}
return a.minspf < b.minspf;
} bool cmplo(lotion a, lotion b)
{
return a.spf < b.spf;
} int main()
{
while(scanf("%d%d", &c, &l) != EOF){
for(int i = ; i < c; i++){
scanf("%d%d", &cow[i].minspf, &cow[i].maxspf);
}
//cout<<cow[0].minspf<<" "<<cow[0].maxspf<<endl;
for(int i = ; i < l; i++){
scanf("%d%d", &lo[i].spf, &lo[i].cover);
}
sort(cow, cow + c, cmpcow);
//cout<<cow[0].minspf<<" "<<cow[0].maxspf<<endl;
sort(lo, lo + l, cmplo); int cnt = ;
for(int i = c - ; i >= ; i--){
for(int j = l - ; j >= ; j--){
if(lo[j].spf < cow[i].minspf)break;
if(lo[j].spf <= cow[i].maxspf && lo[j].cover > && lo[j].spf >= cow[i].minspf){
lo[j].cover--;
cnt++;
break;
}
}
}
printf("%d\n", cnt);
}
return ;
}