【BZOJ 1629】 [Usaco2007 Demo]Cow Acrobats

1629: [Usaco2007 Demo]Cow Acrobats

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 657  Solved: 331
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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 within 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. //有三个头牛，下面三行二个数分别代表其体重及力量 //它们玩叠罗汉的游戏，每个牛的危险值等于它上面的牛的体重总和减去它的力量值，因为它要扛起上面所有的牛嘛. //求所有方案中危险值最大的最小

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

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.

HINT

Source

Silver

贪心。

这道题与【noip2012】国王游戏 几乎一样。

首先要明白对于第i个，放在他上面的第j,k个交换顺序对他没有任何影响。

现在可以只考虑第j,k个：

j放在k的上面，当且仅当 wj-sk<wk-sj  移项得 wj+sj<wk+sk

因此我们只要按照wi+si升序排列，依次放置即可。

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int n;struct data{int x,y,s;}a[50005];bool cmp(data a,data b){return a.s<b.s;}int main(){        scanf("%d",&n);for (int i=1;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y),a[i].s=a[i].x+a[i].y;sort(a+1,a+1+n,cmp);int sum=a[1].x,ans=-a[1].y;for (int i=2;i<=n;i++)ans=max(ans,sum-a[i].y),sum+=a[i].x;cout<<ans<<endl;return 0;}