HDU 4070 + 赤裸裸的贪心~~

　　　　　　　　　　　　　　　　　　　　　　　　　　J - Phage War

　　　　　　　　Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

　　Phage War is a little flash game. In this game, we want infect all cells by the transmission and breed of phages. 
Originally, there is a cell infected by phages and this cell can breed a new phage every second. You should know that only the new born phages can inject other cells.

There are n cells around this cell, numbered from 1 to n. If there are Di phages reaching the i-th cell, the cell would be infected, and the phages journey will cost Ti seconds. To simplify it, we assume these phages will stay in this new cell and they can’t infect other cells. And the new cell cannot breed new phages and infect other cells. 
Can you tell me how much time it costs to infect all cells at least?

Input

In the first line there is an integer T (T <= 50), indicates the number of test cases. 
In each case, the first line contains a integers N (1 <= N <= 10^5). Then there are N lines, each line contain two integers Di, Ti (1<=Di, Ti<=100). 

Output

For each case, output the least time needed in one line.(as shown in the sample output)

Sample Input

2
2
2 1
5 6
2
1 11
3 10

Sample Output

Case 1: 11

Case 2: 14

 

　　这是一道赤裸裸的贪心，，，不知为何，今天的思维非常的差感觉都没什么状态，，和队友讨论了很久这道题，起初我们的贪心的想法几乎完全不同，由于还没有和队友培养好默契，从而没有办法很好地在方向上得到一个统一，不过最终还是被队友AC了，成功地防止了我们爆零的节奏，，，，以此反省自己，作为队长一定要好好兼顾好每个人的想法。。。

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 const int MAXN = ;
 struct node{
     int d,t;
 };
 bool cmp(const node& a,const node& b){
     return a.t > b.t;
 }
 node bug[MAXN];
 int main()
 {
     int t,n,cas = ;
     cin>>t;
     while(t--){
         scanf("%d",&n);
         for(int i = ;i <= n;i++)
             scanf("%d%d",&bug[i].d,&bug[i].t);
         sort(bug + ,bug +  + n,cmp);
         int ans = ,tmp,acum = ;
         for(int i = ;i <= n;i++){
             tmp = bug[i].d + bug[i].t + acum;
             if(ans < tmp) ans = tmp;
             acum += bug[i].d;
         }
         printf("Case %d: %d\n",cas++,ans);
     }
     return ;
 }