POJ3614 [USACO07NOV]防晒霜Sunscreen

Sunscreen

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9333		Accepted: 3264

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 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 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 SPF_i (1 ≤ SPF_i ≤ 1,000). Lotion bottle i can cover cover_i 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: minSPF_i and maxSPF_i
* Lines C+2..C+L+1: Line i+C+1 describes 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 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

USACO 2007 November Gold

 

【题解】

把min从大到小排序，然后对于每个区间，放最大的SPF。分情况讨论两个区间的先后位置可得证，证明显然。

还可用二分图最大匹配做。

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <string>
 #include <algorithm>

 const int INF = 0x7fffffff;
 const int MAXN =  + ; 

 inline void read(int &x)
 {
     x = ;char ch = getchar(),c = ch;
     while(ch < '' || ch > '')c = ch, ch = getchar();
     while(ch <= '' && ch >= '')x = x *  + ch - '', ch = getchar();
     if(c == '-')x = -x;
 }

 int l[MAXN], r[MAXN], cnt[MAXN], point[MAXN], num[MAXN], cntt[MAXN], C, L, ans;

 bool cmp(int a, int b)
 {
     return l[a] > l[b];
 }

 bool cmpp(int a, int b)
 {
     return point[a] > point[b];
 }

 int main()
 {
     //freopen("data.txt", "r", stdin);
     read(C),read(L);
     for(register int i = ;i <= C;++ i) read(l[i]), read(r[i]), cnt[i] = i;
     for(register int i = ;i <= L;++ i) read(point[i]), read(num[i]), cntt[i] = i;
     std::sort(cnt + , cnt +  + C, cmp);
     std::sort(cntt + , cntt +  + L, cmpp);
     for(register int i = ;i <= C;++ i)
     {
         for(register int j = ;j <= L;++ j)
             if(num[cntt[j]] >  && r[cnt[i]] >= point[cntt[j]] && l[cnt[i]] <= point[cntt[j]])
             {
                 --num[cntt[j]], ++ ans;
                 break;
             }
             else if(l[cnt[i]] > point[cntt[j]] ) break;
     }
     printf("%d", ans);
     return ;
 }

POJ3614