BZOJ 1629: [Usaco2007 Demo]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 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

翻译

农民约翰的N（1≤n≤50000）头牛（编号为1—n）计划逃跑去参加马戏团。而他们的蹄子迫使他们不能走钢丝和荡秋千。因此，他们决定练习杂技表演。他们只想出了一个特技动作：站在彼此的头上做“叠罗汉”。

每一个奶牛都有一个重量（1 < = w_i＜＝10000）和承受强度（1 < = s_i < = 1000000000）。一头牛的受伤风险是所有在他上面的牛的总重量（不包括她自己的体重）减去她承受强度（差越小风险越少，差可能为负数）。你的任务是找出所有方案中“牛最大危险值”最小的方案，并输出这个方案的“牛最大危险值”。

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.

第一行为N，接下来N行，每行一对wi和si。

Output

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

输出最小的“牛最大危险值”

Sample Input

3
10 3
2 5
3 3

Sample Output

2

 

题解

贪心。注意：贪心要先写出使状态更优的不等式，并将此转换作为快速排序的准则。

#include<cstdio> #include<cstring> #include<iostream> #include<cstdlib> #include<cmath> #include<algorithm> #define inf 1<<30  using namespace std; int n,sum[50002],ans; struct dian{int x,y;} a[50002]; void init() {     scanf("%d",&n);     int i;     for(i=1;i<=n;i++)        scanf("%d%d",&a[i].x,&a[i].y); } bool kp(const dian &i,const dian &j) {     if(i.x-j.y<j.x-i.y) return true;     return false; } int main() {     init();     sort(a+1,a+n+1,kp);     int i;     ans=-inf;     for(i=1;i<=n;i++)        {if(sum[i-1]-a[i].y>ans) ans=sum[i-1]-a[i].y;         sum[i]=sum[i-1]+a[i].x;            }     printf("%d\n",ans);     return 0; }