2017ACM-北京网络赛-F-HihoCode-1583-Cake

Description

    To celebrate that WF-2018 will be held in PKU, Alice, Bob, and Cate are asked 
to make

    N cakes.

    Every cake i needs to go through 3 steps in restrict order:

    1. Alice mixes flour and water for ai minutes;

    2. Bob carefully bakes it for bi minutes;

    3. Cate makes cream decoration for ci minutes.

    Since Cate wants to have different kinds of cakes, the third step of any cake i
is always not less time-consuming than the second step of any cake j. Also, it is
reasonable that once anyone starts to process a cake, the procedure cannot be
stopped then be resumed.

    To have these cakes done as soon as possible, they need your help.

Input

    There are several cases (less than 15 cases).

    The first line of every case contains an integer N (1 ≤ N ≤ 105)—the number of
cakes to prepare.

    After that, N lines follow, each of them contains three integers ai, bi and ci
(1 ≤ i ≤ N; 0 <ai, bi, ci < 106)—time that needs to be spent on the three steps of
cake i respectively.

    It is guaranteed that for any i and any j, bi is no greater than cj.

    The input ends with N = 0.

Output

For every case, print in a single line the least possible time to make all cakes.

Examples

Input

3
5 3 4
3 2 9
3 4 8
0

Output

26

Problem Description

做蛋糕有三个工序，分别三个人负责，而且必须严格按照步骤来。并且一旦开始做一块蛋糕，这个过程不能停。
否则就得重做。
总共要做n个蛋糕，做第i个蛋糕三个步骤的时间分别为ai，bi，ci.(保证max(bi)<=min(ci) ).

Solution

题解很简单，但是很难想…

$$> ans=max(\sum_i^na_i+min_{1\le i \le n}(b_i+c_i),\sum_i^nc_i+min_{1\le i \le n}(a_i+b_i)) >$$

大致意思就是，所有a的和为suma，所有b的和为sumb，两者最大的一个加上对应剩余两个工序相加的最小值。

Code

/*
 * @Author: Simon 
 * @Date: 2018-08-24 14:46:18 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-08-24 19:15:35
 */
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 100005
int a[maxn];
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    while(cin>>n&&n)
    { 
        int Minxy=INF,Minyz=INF,ans1=0,ans2=0;
        for(int i=1;i<=n;i++)
        {
            int x,y,z;
            cin>>x>>y>>z;
            ans1+=x;
            ans2+=z;
            Minxy=min(x+y,Minxy);
            Minyz=min(y+z,Minyz);
        }
        if(ans1>ans2)
        {
            cout<<ans1+Minyz<<endl;
        }
        else cout<<ans2+Minxy<<endl;
    }
    cin.get(),cin.get();
    return 0;
}