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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7450 | Accepted: 2627 |
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
题意：有C头牛晒太阳，每头牛都有相应的SPF范围，在范围内可以让牛享受太阳，有L瓶防晒霜，每瓶防晒霜都有各自的SPF值，SPF值在某头牛的SPF范围内的防晒霜如果涂在该牛上，可以让该牛享受太阳，问有多少头牛可以享受太阳。
思路：贪心，对于每一种防晒霜，在符合条件的范围内(防晒霜的SPF值大于牛的min_SPF)，每次都筛选出一群牛，将这群牛的max_SPF值压入最小堆，之后尽量用SPF值较小的防晒霜涂于max_SPF较小的牛上。
#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<queue> #include<algorithm> #include<functional>//greater using namespace std; const int N_MAX = 2500; struct cows { int low; int high; bool operator <(const cows&b)const{ return low<b.low||(low==b.low&&high<b.high); } }; struct lotion { int spf; int num; bool operator <(const lotion&b)const { return spf <b.spf ; } }; priority_queue<int,vector<int>,greater<int>>que;//小元素在上 cows cow[N_MAX]; lotion lo[N_MAX]; int main() { int C, L; scanf("%d%d",&C,&L); for (int i = 0;i < C;i++) scanf("%d%d", &cow[i].low, &cow[i].high); for (int i = 0;i < L;i++) scanf("%d%d", &lo[i].spf, &lo[i].num); sort(cow, cow + C);//按牛的spf最小值从小到大排序 sort(lo, lo + L);//按护肤品的spf值从小到大排 int result=0,cur=0; for (int i = 0;i < L;i++) {//对于每一种护肤品进行考虑 while (cur<C&&lo[i].spf>=cow[cur].low) {//挑选出满足条件的牛的high值进堆 que.push(cow[cur].high);//进堆后小的high值在上面 cur++; } while (que.size() && lo[i].num) {//如果这种护肤品对所有挑选出来的牛都不满足条件，那么只能弃用，因为后面的牛的min值都比护肤品的spf大 int k = que.top();que.pop(); if (k >= lo[i].spf) {//high值小于护肤品的spf值的牛只能不涂，后面的护肤品spf值越来越大更不满足 result++; lo[i].num--; } } } printf("%d\n",result); return 0; }
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原文地址：http://www.cnblogs.com/ZefengYao/p/5893893.html
