sgu259 Printed PR 贪心

link:http://acm.sgu.ru/problem.php?contest=0&problem=259

思路就是贪心．

首先要读懂题目，输入的方式，把样例读懂．

第一，打印的总时间一定．需要做的就是送出的时间尽可能的重合，这样总时间就会更少．所以，送出时间长的要尽可能的先打印，按照送出时间从大到小排序就可以了．

 /*
  * =====================================================================================
  *       Filename:  259.cpp
  *        Created:  06/08/2013 10:15:09
  *         Author:  liuxueyang (lxy), 1459917536@qq.com
  *   Organization:  Hunan University
  *
  * =====================================================================================
  */

 /*
 ID: zypz4571
 LANG: C++
 TASK: 259.cpp
 */
 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cctype>
 #include <algorithm>
 #include <queue>
 #include <set>
 #include <stack>
 #include <map>
 #include <list>
 #define INF 0x3f3f3f3f
 #define MOD 1000000007
 #define LL long long
 const double eps=1e-;
 using namespace std;
 struct Node{
     int t, l;
     bool operator < (const Node other) const {
         if (l!=other.l) return l>other.l;
         else return t<other.t;
     }
 }a[];
 int main ( int argc, char *argv[] )
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
 #endif
     ios::sync_with_stdio(false);
     int n;
     while (cin>>n) {
         for (int i = ; i < n; ++i) cin>>a[i].t;
         for (int i = ; i < n; ++i) cin>>a[i].l;
         int ans=, sum=;
         sort(a,a+n);
         for (int i = ; i < n; ++i)  {
             sum += a[i].t;
             ans = max(ans, sum+a[i].l);
         }
         cout<<ans<<endl;
     }
         return EXIT_SUCCESS;
 }                /* ----------  end of function main  ---------- */

囧

开始要读懂题目意思啊，读题认真．