POJ 3045 贪心

Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join the circus. Their hoofed feet prevent them from tightrope walking and swinging from the trapeze (and their last attempt at firing a cow out of a cannon met with a dismal failure). Thus, they have decided to practice performing acrobatic stunts.

The cows aren't terribly creative and have only come up with one acrobatic stunt: standing on top of each other to form a vertical stack of some height. The cows are trying to figure out the order in which they should arrange themselves ithin this stack.

Each of the N cows has an associated weight (1 <= W_i <= 10,000) and strength (1 <= S_i <= 1,000,000,000). The risk of a cow collapsing is equal to the combined weight of all cows on top of her (not including her own weight, of course) minus her strength (so that a stronger cow has a lower risk). Your task is to determine an ordering of the cows that minimizes the greatest risk of collapse for any of the cows.Input

* Line 1: A single line with the integer N.

* Lines 2..N+1: Line i+1 describes cow i with two space-separated integers, W_i and S_i.

Output

* Line 1: A single integer, giving the largest risk of all the cows in any optimal ordering that minimizes the risk.

Sample Input

3
10 3
2 5
3 3

Sample Output

2

Hint

OUTPUT DETAILS:

Put the cow with weight 10 on the bottom. She will carry the other two cows, so the risk of her collapsing is 2+3-3=2. The other cows have lower risk of collapsing.

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二分也能做的样子，不过感觉有点多此一举了？？……明明贪心可以很好的解决了的……（感谢大佬们的题解……果然现在自己想的话还是比较困难）

大概就是要求最优方案的最大risk，也就是最小化最大值的感觉，每头牛的risk=上面的w之和-它的s,换一个表达方式的话就是risk=从最顶层到当前层的牛的w之和-这头牛的w-这头牛的s

从两头牛考虑，无非一个在下一个在上，那么其中一种方案，W-（wi+si）<W-(wj+sj)，最优方案的选取只需要按wi+si降序排序。

然后选取最优方案中risk的最大值就可以了……

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<string.h>
#include<algorithm>
typedef long long ll;
using namespace std;
struct cows { ll w, s,d; };
int n, k;
cows c[50005];
ll res[50005];
bool cmp(cows a, cows b) { return a.d > b.d; };
int main()
{
    while (scanf("%lld", &n) != EOF)
    {
        memset(res, 0, sizeof(res));
        for (int i = 0; i < n; i++)
        {
            scanf("%lld%lld", &c[i].w, &c[i].s);
            c[i].d = c[i].w + c[i].s;
        }
        sort(c, c + n,cmp);
        ll maxx = -0x3f3f3f3f;
        res[n - 1] = 0;
        for (int i = n-2; i>=0; i--)
        {
            res[i] += res[i + 1] + c[i+1].w;
        }
        for (int i = 0; i < n; i++)
        {
            maxx = max(res[i] - c[i].s,maxx);
        }
        printf("%lld\n", maxx);
    }
    return 0;
}

 

 

 

转载于:https://www.cnblogs.com/Egoist-/p/7406585.html