poj3045 Cow Acrobats(二分最大化最小值)

https://vjudge.net/problem/POJ-3045

读题后提取到一点：例如对最底层的牛来说，它的崩溃风险=所有牛的重量-(底层牛的w+s)，则w+s越大，越在底层。

注意范围lb=-INF。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<map>
 8 #define lson l, m, rt<<1
 9 #define rson m+1, r, rt<<1|1
10 #define INF 0x3f3f3f3f
11 typedef long long ll;
12 using namespace std;
13 typedef struct{
14     ll w, s;
15     ll sum;
16 }Node;
17 Node node[100010];
18 ll n, sum=0;
19 int C(int x)
20 {
21     ll maxm = -INF, _sum=sum;
22     for(int i = 0; i < n; i++){
23         ll tmp = _sum-node[i].sum;
24         _sum -= node[i].w;
25         if(tmp>x) return 0; 
26     }
27     return 1;
28 }
29 bool cmp(const Node a, const Node b)
30 {
31     if(a.sum != b.sum)
32         return a.sum>b.sum;
33     else return a.s>b.s;
34 } 
35 int main()
36 {
37     scanf("%lld", &n);
38     for(int i = 0; i < n; i++){
39         scanf("%lld%lld", &node[i].w, &node[i].s);
40         node[i].sum = node[i].w+node[i].s;
41         sum += node[i].w; 
42     }
43     sort(node, node+n, cmp);
44     ll lb=-INF, ub=INF;
45     while(ub-lb>1){
46         ll mid = (lb+ub)>>1;
47         if(C(mid)){
48             ub = mid;
49         }
50         else lb = mid;
51     }
52     printf("%lld\n", ub);
53     return 0;
54 }