Codeforces 230C Shifts [생각]

C. Shifts
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a table consisting of n rows and m columns. Each cell of the table contains a number, 0 or 1. In one move we can choose some row of the table and cyclically shift its values either one cell to the left, or one cell to the right.
To cyclically shift a table row one cell to the right means to move the value of each cell, except for the last one, to the right neighboring cell, and to move the value of the last cell to the first cell. A cyclical shift of a row to the left is performed similarly, but in the other direction. For example, if we cyclically shift a row "00110"one cell to the right, we get a row "00011", but if we shift a row "00110"one cell to the left, we get a row "01100".
Determine the minimum number of moves needed to make some table column consist only of numbers 1.
Input
The first line contains two space-separated integers: n (1 ≤ n ≤ 100) — the number of rows in the table and m (1 ≤ m ≤ 104) — the number of columns in the table. Then n lines follow, each of them contains m characters "0"or "1": the j-th character of the i-th line describes the contents of the cell in the i-th row and in the j-th column of the table.
It is guaranteed that the description of the table contains no other characters besides "0"and "1".
Output
Print a single number: the minimum number of moves needed to get only numbers 1 in some column of the table. If this is impossible, print -1.
Examples
input

3 6
101010
000100
100000

output

3

input

2 3
111
000

output

-1

Note
In the first sample one way to achieve the goal with the least number of moves is as follows: cyclically shift the second row to the right once, then shift the third row to the left twice. Then the table column before the last one will contain only 1s.
In the second sample one can't shift the rows to get a column containing only 1s.
제목 대의:
줄마다 순환하여 위치를 옮길 수 있다. 최소한 몇 번을 옮기면 어떤 열이 모두 1이 될 수 있느냐고 묻는다.
아이디어:
① 우리는 L[i][j]를 미리 처리하여 i행은 위치 j에서 가장 왼쪽, j점에서 가장 가까운 1의 위치를 나타낸다.
같은 도리로 R[i][j]가 있다.
② 그리고 우리는 LL[i]가 i행의 맨 왼쪽에 있는 1의 자리를 미리 처리한다.
같은 이치로 RR【i】이 있다.
③ 우리 O(m)는 일렬로 열거하여 최종 일렬이 모두 1임을 나타낸다. 그리고 어느 1이 가장 좋은지 욕심을 내어 판단하면 된다.
프로세스 유지 보수 좀 해주세요.
Ac 코드:

#include
#include
#include
#include
using namespace std;
char a[150][15000];
int L[150][15000];
int R[150][15000];
int LL[150];
int RR[150];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(LL,-1,sizeof(LL));
        memset(RR,-1,sizeof(RR));
        memset(L,-1,sizeof(L));
        memset(R,-1,sizeof(R));
        for(int i=1;i<=n;i++)scanf("%s",a[i]+1);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(LL[i]==-1&&a[i][j]=='1')LL[i]=j;
                if(a[i][j]=='1')L[i][j]=j;
                else L[i][j]=L[i][j-1];
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=1;j--)
            {
                if(RR[i]==-1&&a[i][j]=='1')RR[i]=j;
                if(a[i][j]=='1')R[i][j]=j;
                else R[i][j]=R[i][j+1];
            }
        }
        int output=0x3f3f3f3f;
        for(int i=1;i<=m;i++)
        {
            int sum=0;
            for(int j=1;j<=n;j++)
            {
                int A=15000000;
                int B=15000000;
                if(L[j][i]!=-1)A=i-L[j][i];
                if(R[j][i]!=-1)B=R[j][i]-i;
                int C=15000000;
                int D=15000000;
                if(LL[j]!=-1)C=LL[j]+m-i;
                if(RR[j]!=-1)D=m-RR[j]+i;
                sum+=min(min(A,B),min(C,D));
            }
            output=min(output,sum);
        }
        if(output>=15000000)printf("-1
");
        else printf("%d
",output);
    }
}