首页 > > 程序设计 > C/C++ >

Gym 100703I---Endeavor for perfection（尺取）

2018-06-17 23:40:55来源：未知阅读 ()

题目链接

http://codeforces.com/problemset/gymProblem/100703/I

Description

standard input/output
Statements

As a matter of fact, Dragon knows what Prince is interested in now. Prince uses to spend his rare off days learning different courses and trainings. But Dragon doubts whether he should tell Princess about it.

Prince decided that he needs some extra knowledge and skills. He chose n fields in which he wanted to gain knowledge and skills most of all. After this, he learned that m₁, m₂, ..., m_n courses and trainings of each of fields exist.

Prince took a close look at descriptions of courses and trainings and drew a table, in which s_ij is a value by which ith skill is increased after studying jth course (j = 1, 2, ..., m_i).

Prince believes that his basic knowledge and skills in all these fields are negligible, so they can be considered zero. He wants to evolve his knowledge and skills harmonically. In his opinion, he will reach the greatest harmony if he chooses one course for each field in such a way that difference between the highest and the lowest their increases would be as minimum as possible.

Your task is to find the courses which Prince should choose.

Input

The first line contains integer n (1 ≤ n ≤ 200) — the number of fields which Prince is interested in.

The second line contains n integers m₁, m₂, ..., m_n (1 ≤ m_j ≤ 1000, j = 1, 2, ..., n) — the number of courses for each of fields.

The next n lines contain values s_ij (1 ≤ s_ij ≤ 10⁹) — knowledges and skills, which Prince would gain at the courses. The first of thesen lines contains values s₁₁, s₁₂, ..., s_1m₁, the second — values s₂₁, s₂₂, ..., s_2m₂, etc.

The values s_ij are listed in the numerical order of courses for each of the fields.

Output

In the first line print one integer — minimum difference between the highest and the lowest numbers of increase.

In the second line print n integers — numbers of courses which Prince should choose. List the numbers in the same order in which the fields are listed.

If there is more than one answer — choose any of them.

Sample Input

Input

Output

0
2 1

Input

Output

3
2 5 4 4 


题意：输入一个n，然后输入n个数，表示接下来输入的n行每行的数的个数，求在每行中选择一个数使得这n个数的最大值与最小值的差最小，输出最小的差值和每行选择的数的列号；

思路：尺取，将n行的数放在一起从小到大排序，定义s=0和e=0，表示s~e的一段区间，e向右移动，直到这个区间包含n行的数，那么node[e].x-node[s].x便是从这个区间选择的n行数的最小差值，然后s++，再让e右移，计算区间n行数最小差值......

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <bitset>
using namespace std;
const int M=1e9+5;
const int maxn=2e5+5;
int A[205],cnt[205],vis[205];
struct Node
{
    int x,h,l;
}node[maxn],ans[205];
bool cmp1(const Node s1,const Node s2)
{
    return s1.x<s2.x;
}
bool cmp2(const Node s1,const Node s2)
{
    return s1.h<s2.h;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
       memset(cnt,0,sizeof(cnt));
       memset(vis,0,sizeof(vis));
       for(int i=0;i<n;i++)
           scanf("%d",&A[i]);
       int tot=0;
       for(int i=0;i<n;i++)
       for(int j=1;j<=A[i];j++)
       {
           scanf("%d",&node[tot].x);
           node[tot].h=i;
           node[tot++].l=j;
       }
       sort(node,node+tot,cmp1);
       int tmp=M,s=0,e=0,sum=0;
       int pos1,pos2;
       while(1)
       {
           while(e<tot&&sum<n){
               cnt[node[e].h]++;
               if(cnt[node[e].h]==1) sum++;
               e++;
           }
           if(sum<n) break;
           if(node[e-1].x-node[s].x<tmp){
               tmp=node[e-1].x-node[s].x;
               pos1=s;
               pos2=e-1;
           }
           if(tmp==0) break;
           if(cnt[node[s].h]==1) sum--;
           cnt[node[s].h]--;
           s++;
       }
       int p=0;
       for(int i=pos1;i<=pos2;i++)
       {
           if(vis[node[i].h]==0)
           {
              vis[node[i].h]=1;
              ans[p].h=node[i].h;
              ans[p++].l=node[i].l;
           }
       }
       sort(ans,ans+n,cmp2);
       printf("%d\n",tmp);
       for(int i=0;i<n;i++)
        printf("%d%c",ans[i].l,(i+1==n)?'\n':' ');
    }
    return 0;
}