http://poj.org/problem?id=3622
貪心,即排序後對於每頭牛選擇能滿足他的最便宜的草,注意要從最挑剔的牛(即greenness最大的牛)開始,但是O(nm)會超時。O(nlogm)的做法跟O(nm)的一樣,只是在選擇草的時候用二分。
代碼中因爲藉助了multiset,所以排序的時候只需按照greenness排序。
代碼:
#include <cstdio>
#include <set>
#include <algorithm>
using namespace std;
const int N = 100000 + 10;
struct NODE
{
int p, g;
}cow[N], grass[N];
multiset<int> ms;
bool cmp(const NODE &a, const NODE &b) //because of the ms, we needn't to sort by price
{
return a.g <= b.g;
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++)
scanf("%d%d", &cow[i].p, &cow[i].g);
for (int i = 0; i < m; i++)
scanf("%d%d", &grass[i].p, &grass[i].g);
sort(cow, cow+n, cmp);
sort(grass, grass+m, cmp);
int j = m - 1;
long long sum = 0;
bool flag = true;
ms.clear();
for (int i = n-1; i >= 0; i--)
{
while (j >= 0 && grass[j].g >= cow[i].g)
ms.insert(grass[j--].p);
multiset<int>::iterator it = ms.lower_bound(cow[i].p);
if (it == ms.end())
{
flag = false;
break;
}
else
{
sum += *it;
ms.erase(it);
}
}
printf("%lld\n", flag ? sum : -1);
}