[bzoj1629][Usaco2007 Demo]Cow Acrobats

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探讨了如何通过算法确定牛叠罗汉游戏的最佳顺序，以最小化最危险情况的风险值。采用排序与遍历的方法，实现了计算最优解的目标。

1629: [Usaco2007 Demo]Cow Acrobats

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 975 Solved: 507
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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

10 3

2 5

3 3

Sample Output

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

sol:
如果2只奶牛交换位置。设奶牛的力量为s,重量为w,设1号奶牛更优秀，则有w1+sum-s2 < w2+sum-s1(另一牛在上方),整理w1+s1 < w2+s2，则总和小的更加优秀。

#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
using namespace std;
int n,m;
inline int read()
{
    char c;
    int res,flag=0;
    while((c=getchar())>'9'||c<'0') if(c=='-')flag=1;
    res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=(res<<3)+(res<<1)+c-'0';
    return flag?-res:res;
}
const int N=1000000;
struct cc
{
    int x,y;
}a[N];
inline bool cmp(const cc &a,const cc &b)
{
    return a.x+a.y<b.x+b.y;
}
int main()
{
//  freopen("1629.in","r",stdin);
//  freopen(".out","w",stdout);
    n=read();
    for(int i=1;i<=n;++i)
    a[i].x=read(),a[i].y=read();
    sort(a+1,a+1+n,cmp);
    int sum=0,ans=-1e9;
    for(int i=1;i<=n;++i)
    {
        ans=max(ans,sum-a[i].y);
        sum+=a[i].x;
    }
    printf("%d",ans);
}