BZOJ1629: [Usaco2007 Demo]Cow Acrobats

1629: [Usaco2007 Demo]Cow Acrobats

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 601  Solved: 305
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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

题解:
同国王游戏,只不是按a[i]+b[i]递增排序。。。
正确性可以从 任意交换两个a[i]+b[i]逆序的,答案不会变差来考虑
WA了一次是因为危险值竟然还有负的。。。
代码:
 
 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 50000+100
14 #define maxm 500+100
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 using namespace std;
19 inline int read()
20 {
21     int x=0,f=1;char ch=getchar();
22     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
23     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
24     return x*f;
25 }
26 int n;
27 struct rec{int x,y;}a[maxn];
28 inline bool cmp(rec a,rec b)
29 {
30     return a.x+a.y<b.x+b.y;
31 }
32 int main()
33 {
34     freopen("input.txt","r",stdin);
35     freopen("output.txt","w",stdout);
36     n=read();
37     for(int i=1;i<=n;i++)a[i].x=read(),a[i].y=read();
38     sort(a+1,a+n+1,cmp);
39     int sum=0,ans=-inf;
40     for(int i=1;i<=n;i++)
41     {
42         ans=max(ans,sum-a[i].y);
43         sum+=a[i].x;
44     }
45     printf("%d\n",ans);
46     return 0;
47 }
View Code

 

posted @ 2014-08-28 18:38  ZYF-ZYF  Views(149)  Comments(0Edit  收藏  举报