POJ_3045_Cow_Acrobats

描述

分析

易证:w+s越大应在越下面.所以排序,扫一遍即可.

注意:

1.ans的初始值应为-INF而非0,因为很可能大家的力气都很大,但都很轻!

 1 #include<cstdio>
 2 #include<algorithm>
 3 using std :: sort;
 4 using std :: max;
 5
 6 const int maxn=50005,INF=0x7fffffff;
 7 int n;
 8
 9 struct point
10 {
11     int w,s,sum;
12 }c[maxn];
13
14 bool comp(point x,point y) { return x.sum<y.sum; }
15
16 void solve()
17 {
18     sort(c+1,c+n+1,comp);
19     int ans=-INF,sum=0;
20     for(int i=1;i<=n;i++)
21     {
22         ans=max(ans,sum-c[i].s);
23         sum+=c[i].w;
24     }
25     printf("%d\n",ans);
26 }
27
28 void init()
29 {
30     scanf("%d",&n);
31     for(int i=1;i<=n;i++)
32     {
33         scanf("%d%d",&c[i].w,&c[i].s);
34         c[i].sum=c[i].w+c[i].s;
35     }
36 }
37
38 int main()
39 {
40     freopen("cow.in","r",stdin);
41     freopen("cow.out","w",stdout);
42     init();
43     solve();
44     fclose(stdin);
45     fclose(stdout);
46     return 0;
47 }