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 = 2500 + 10; 

inline void read(int &x)
{
    x = 0;char ch = getchar(),c = ch;
    while(ch < '0' || ch > '9')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', 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 = 1;i <= C;++ i) read(l[i]), read(r[i]), cnt[i] = i;
    for(register int i = 1;i <= L;++ i) read(point[i]), read(num[i]), cntt[i] = i;
    std::sort(cnt + 1, cnt + 1 + C, cmp);
    std::sort(cntt + 1, cntt + 1 + L, cmpp);
    for(register int i = 1;i <= C;++ i)
    {
        for(register int j = 1;j <= L;++ j)
            if(num[cntt[j]] > 0 && 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 0;
}