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。

 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 #define INF 0x3f3f3f3f
 typedef long long ll;
 using namespace std;
 typedef struct{
     ll w, s;
     ll sum;
 }Node;
 Node node[];
 ll n, sum=;
 int C(int x)
 {
     ll maxm = -INF, _sum=sum;
     for(int i = ; i < n; i++){
         ll tmp = _sum-node[i].sum;
         _sum -= node[i].w;
         if(tmp>x) return ;
     }
     return ;
 }
 bool cmp(const Node a, const Node b)
 {
     if(a.sum != b.sum)
         return a.sum>b.sum;
     else return a.s>b.s;
 }
 int main()
 {
     scanf("%lld", &n);
     for(int i = ; i < n; i++){
         scanf("%lld%lld", &node[i].w, &node[i].s);
         node[i].sum = node[i].w+node[i].s;
         sum += node[i].w;
     }
     sort(node, node+n, cmp);
     ll lb=-INF, ub=INF;
     while(ub-lb>){
         ll mid = (lb+ub)>>;
         if(C(mid)){
             ub = mid;
         }
         else lb = mid;
     }
     printf("%lld\n", ub);
     return ;
 }