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Cow Acrobats
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 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 |
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题目的意思是给出每个人的力量值和体重，现在把每个人垒起来，每个人的压力等于他上面的人的总质量减去他的力量，求怎么排最大危险系数最小
思路：贪心把力量和体重之和大的排下面
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <vector>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long long int n;
struct node{
int w,s;
}p[100005]; bool cmp(node a,node b)
{
return a.w+a.s<b.w+b.s;
} int main()
{
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
scanf("%d%d",&p[i].w,&p[i].s);
sort(p,p+n,cmp);
int mx=-p[0].s;
int sum=0;
for(int i=0;i<n-1;i++)
{
sum+=p[i].w;
mx=max(mx,sum-p[i+1].s);
}
printf("%d\n",mx);
} return 0;
}