POJ 3614 Sunscree （优先队列）

Sunscreen

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 10586	Accepted: 3702

Description

To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide withsunscreen when they're at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 1,000) that will work. If the SPF rating is too low, the cow sufferssunburn; if the SPF rating is too high, the cow doesn'ttan at all........

The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion,each bottle i with an SPF rating SPF_i (1 ≤ SPF_i ≤ 1,000). Lotion bottle i can cover cover_i cows with lotion. A cow may lotionfrom only one bottle.

What is the maximum number of cows that can protect themselveswhile 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 twointegers: minSPF_i and maxSPF_i 
* Lines C+2..C+L+1:Line i+C+1describes a sunscreen lotion bottle i with space-separated integers: SPF_i and cover_i

Output

A single line with an integer that is the maximum number of cowsthat can be protected while tanning

SampleInput

3 2

3 10

2 5

1 5

6 2

4 1

SampleOutput

2

Source

​​USACO 2007 November Gold​​

 

算法分析：题意：n个母牛要抹上防晒物品，对防晒物品有系数要求，，给出若干种防晒品的防晒系数和数量，问最多多少奶牛能使用防晒物品。

分析：

牛按最小值升序排序，物品按系数升序排序。维护一个牛最大程度的优先队列，选取规则为最小程度小于当前物品的系数，然后优先对列出队，是否可以选择物品。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
#define N 2550
struct node
{
  int x,y;
}cow[N],l[N];

int n,m; 
bool cmp(const node&a,const node&b)  
{ 
  
  return  a.x<b.x;
  
}  
int solve()
{
  sort(cow+1,cow+n+1,cmp);
  sort(l+1,l+m+1,cmp);
  priority_queue<int,vector<int>,greater<int> >q;//从小到大牌排列，默认是从大到小 
  int ans=0,cur=1;
  for(int i=1;i<=m;i++)
  {
    while(cur<=n&&cow[cur].x<=l[i].x)
    {
        
      q.push(cow[cur].y);  //母牛进队
      ++cur;               //进队母牛+1
    }
    while(!q.empty()&&l[i].y)
    {
      int t=q.top();
      q.pop(); 
      
      if(t>=l[i].x)
      {
         ans++;      //可以使用
         --l[i].y;   //物品数量-1
      }
    }
  }
  return ans;
}
int main()
{
  
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&cow[i].x,&cow[i].y);
    }
    for(int i=1;i<=m;i++)
    {
      scanf("%d%d",&l[i].x,&l[i].y);
    }
    printf("%d\n",solve());
  }
  return 0;
  
}