P2887 [USACO07NOV]防晒霜Sunscreen
题目描述
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?
有C个奶牛去晒太阳 (1 <=C <= 2500),每个奶牛各自能够忍受的阳光强度有一个最小值和一个最大值,太大就晒伤了,太小奶牛没感觉。
而刚开始的阳光的强度非常大,奶牛都承受不住,然后奶牛就得涂抹防晒霜,防晒霜的作用是让阳光照在身上的阳光强度固定为某个值。
那么为了不让奶牛烫伤,又不会没有效果。
给出了L种防晒霜。每种的数量和固定的阳光强度也给出来了
每个奶牛只能抹一瓶防晒霜,最后问能够享受晒太阳的奶牛有几个。
输入格式
* 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
输出格式
A single line with an integer that is the maximum number of cows that can be protected while tanning
输入输出样例
3 2 3 10 2 5 1 5 6 2 4 1
2
区间覆盖问题,求线段最多覆盖多少点,线段按右端点排序,防晒霜按防护值排序
之后进行贪心即可,注意防晒霜的数量
代码:
#include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int N=5010; int n,m; int ans; struct cow { int l,r; } a[N]; struct sun { int n,s; } b[N]; bool cmp1(cow x,cow y) { if(x.r<y.r) return true; return false; } bool cmp2(sun a,sun b) { if(a.n<b.n) return true; return false; } int main () { scanf("%d%d",&n,&m); for(int i=1; i<=n; i++) scanf("%d%d",&a[i].l,&a[i].r); for(int i=1; i<=m; i++) scanf("%d%d",&b[i].n,&b[i].s); sort(a+1,a+n+1,cmp1); sort(b+1,b+m+1,cmp2); for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++) if(b[j].n>=a[i].l&&b[j].n<=a[i].r&&b[j].s) { b[j].s--; ans++; break; } } printf("%d\n",ans); return 0; }