POJ_3045_Cow_Acrobats_(贪心)

描述

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http://poj.org/problem?id=3045

n头牛,每头牛都有重量w[i]和力量s[i].把这n头牛落起来,每头牛会有一个危险值,危险值是它上面所有牛的重量和减去它的力量.求危险值最大的牛的危险值的最小值.

Cow Acrobats

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4102		Accepted: 1569

Description

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.

Source

USACO 2005 November Silver

分析

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贪心.

易证:w+s越大应在越下面(又重,又有力气,当然放在下面...).

所以排序,扫一遍即可.

 

注意:

1.ans的初始值应为-INF而非0,因为很可能大家的力气都很大,但都很轻! 

#include<cstdio>
#include<algorithm>
using std :: sort;
using std :: max;

const int maxn=50005,INF=0x7fffffff;
int n;

struct point
{
    int w,s,sum;
}c[maxn];

bool comp(point x,point y) { return x.sum<y.sum; }

void solve()
{
    sort(c+1,c+n+1,comp);
    int ans=-INF,sum=0;
    for(int i=1;i<=n;i++)
    {
        ans=max(ans,sum-c[i].s);
        sum+=c[i].w;
    }
    printf("%d
",ans);
}

void init()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    { 
        scanf("%d%d",&c[i].w,&c[i].s); 
        c[i].sum=c[i].w+c[i].s; 
    }
}

int main()
{
    freopen("cow.in","r",stdin);
    freopen("cow.out","w",stdout);
    init();
    solve();
    fclose(stdin);
    fclose(stdout);
    return 0;
}