hdu 5092 Seam Carving
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分析:一道简单的dp题,就是找到从上到下每行取一个点,使之联通且和最小,联通就是指 8个方向联通。
dp[i][j] 表示从下到上 以 i,j为终点的最小的联通线。从最后一行到最上面一行dp。然后找到最小的值,从上往下找如果
dp[i][j] == dp[i+1][k] + a[i][j]( k = j-1, j, j+1)则往这个方向走,记得从右往左找。因为相同大小取最右边。
代码:
#include<cstdio>#include<iostream>using namespace std;const int N = 105;const int INF = 1<<30;int a[N][N];int dp[N][N];int main(){ // freopen("5092.txt", "r", stdin); int T, m, n; scanf("%d", &T); for(int cas = 1; cas <= T; cas++){ printf("Case %d\n", cas); scanf("%d%d", &n, &m); for(int i = 0; i <= n+1; i++) dp[i][0] = dp[i][m+1] = INF; for(int i = 0; i <= m+1; i++) dp[0][i] = dp[n+1][i] = INF; for(int i = 1; i <= n; i++){ for(int j = 1; j <= m; j++){ scanf("%d", &a[i][j]); } } for(int i = 1; i <= m; i++) dp[n][i] = a[n][i]; for(int i = n-1; i >= 1; i--){ for(int j = 1; j <= m; j++){ dp[i][j] = min(min(dp[i+1][j-1],dp[i+1][j]), dp[i+1][j+1]) + a[i][j]; } } int index = 1; for(int i = 2; i <= m; i++){ if(dp[1][i] <= dp[1][index]) index = i; } for(int i = 1; i <= n; i++){ printf("%d%c", index, " \n"[i==n]); if(i == n) break; for(int j = 1; j >= -1; j--){ if(dp[i][index] - a[i][index] == dp[i+1][index+j]){ index += j; break; } } } } return 0;}
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