[DP] poj2353

/* 题意是 有N层，每层M个房间，每个房间有花费， 从第1层走向第N层，可以向上、左、右方向走。 输出使得花费最小的路径 */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int inque[600];
int a[120][600], f[120][600];
int pre[120 * 600], n, m;
void
dfs(int k)
{
  if (k == -1)
    ;
  else
    dfs(pre[k]), cout << k % 501 << endl;
  return;
}
int
main()
{ // freopen("in.txt","r",stdin);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", &a[i][j]);
      if (i == 1)
        f[i][j] = a[i][j], pre[i * 501 + j] = -1;
      else
        f[i][j] = f[i - 1][j] + a[i][j], pre[i * 501 + j] = (i - 1) * 501 + j;
    }
    for (int j = 2; j <= m; j++)
      if (f[i][j] > f[i][j - 1] + a[i][j])
        f[i][j] = f[i][j - 1] + a[i][j], pre[i * 501 + j] = i * 501 + j - 1;
    for (int j = m - 1; j >= 1; j--)
      if (f[i][j] > f[i][j + 1] + a[i][j])
        f[i][j] = f[i][j + 1] + a[i][j], pre[i * 501 + j] = i * 501 + j + 1;
  } 
/* for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) cout<<pre[i*501+j]<<" ";
       //cout<<f[i][j]<<" "; cout<<endl; }*/
  int mark = 0, Min = f[n][1];
  for (int i = 1; i <= m; i++) {
    if (f[n][i] <= Min)
      Min = f[n][mark = i];
  }
  // printf("mark=%d/n",mark);
  dfs(n * 501 + mark);
}