|
Language:
Sunscreen
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 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头牛晒太阳，每头牛都有一个能承受辐射的范围（min~max），现在有 l 种防晒霜，每种防晒霜都能将辐射值固定在spf，每种防晒霜都有一定的数量num。每头牛用最多一种防晒霜，问能满足多少斗牛。
思路：贪心，首先防晒霜按照spf从小到大排序，牛也按照能承受的最小值从小到大排序。然后对于每种防晒霜 将牛的最小值 小于等于 该种防晒霜spf值的牛入队列（只如它承受范围的最大值），然后较小者先出队列。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 3005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct Cow
{
int l,r;
}cow[maxn];
struct Sunscreen
{
int p,num;
}s[maxn];
int c,l;
int cmp1(Cow a,Cow b)
{
return a.l<b.l;
}
int cmp2(Sunscreen a,Sunscreen b)
{
return a.p<b.p;
}
priority_queue<int,vector<int>,greater<int> >Q;
int main()
{
int i,j,t;
while (~sff(c,l))
{
FRL(i,0,c)
sff(cow[i].l,cow[i].r);
FRL(i,0,l)
sff(s[i].p,s[i].num);
sort(cow,cow+c,cmp1);
sort(s,s+l,cmp2);
int ans=0;
i=j=0;
FRL(j,0,l)
{
while (i<c && cow[i].l<=s[j].p)
{
Q.push(cow[i].r);
i++;
}
while (!Q.empty()&&s[j].num)
{
t=Q.top();
Q.pop();
if (s[j].p<=t)
{
ans++;
s[j].num--;
}
}
}
pf("%d\n",ans);
}
return 0;
}
原文地址：http://blog.csdn.net/u014422052/article/details/44467781
