hdoj 5092 Seam Carving 【树塔DP变形 + 路径输出】 【简单题】
Seam Carving
and stored them in his computer. He knew it is an effective way to carve the seams of the images He only knew that there is optical energy in every pixel. He learns the following principle of seam carving. Here seam carving refers to delete through horizontal
or vertical line of pixels across the whole image to achieve image scaling effect. In order to maintain the characteristics of the image pixels to delete the importance of the image lines must be weakest. The importance of the pixel lines is determined in
accordance with the type of scene images of different energy content. That is, the place with the more energy and the richer texture of the image should be retained. So the horizontal and vertical lines having the lowest energy are the object of inspection.
By constantly deleting the low-energy line it can repair the image as the original scene.
For an original image G of m*n, where m and n are the row and column of the image respectively. Fish obtained the corresponding energy matrix A. He knew every time a seam with the lowest energy should be carved. That is, the line with the lowest sum of energy
passing through the pixels along the line, which is a 8-connected path vertically or horizontally.
Here your task is to carve a pixel from the first row to the final row along the seam. We call such seam a vertical seam.
Then on the next m line, there n integers.
than one such seams, just print the column number of the rightmost seam.
2
4 3
55 32 75
17 69 73
54 81 63
47 5 45
6 6
51 57 49 65 50 74
33 16 62 68 48 61
2 49 76 33 32 78
23 68 62 37 69 39
68 59 77 77 96 59
31 88 63 79 32 34
Case 1
2 1 1 2
Case 2
3 2 1 1 2 1
没办法,仅仅好主修图论 + DP。 其他仅仅能辅修了o(╯□╰)o
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct Node
{
int pos;
};
Node mark[110][110];//标记该位置是由 哪一位置得到的(自底往上)
int dp[110][110];
int N, M;
int k = 1;
void getMap()
{
for(int i = 1; i <= N; i++)
{
for(int j = 1; j <= M; j++)
scanf("%d", &dp[i][j]);
}
}
void solve()
{
int t;
for(int i = N-1; i >= 1; i--)
{
for(int j = 1; j <= M; j++)
{
if(j == 1)
{
t = min(dp[i+1][j], dp[i+1][j+1]);//找到最小值
dp[i][j] += t;
//从左到右更新 前驱
if(t == dp[i+1][j])
mark[i][j].pos = j;
if(t == dp[i+1][j+1])
mark[i][j].pos = j + 1;
}
else if(j == M)
{
t = min(dp[i+1][j], dp[i+1][j-1]);
dp[i][j] += t;
if(t == dp[i+1][j-1])
mark[i][j].pos = j - 1;
if(t == dp[i+1][j])
mark[i][j].pos = j;
}
else
{
int t = min(dp[i+1][j], min(dp[i+1][j-1], dp[i+1][j+1]));
dp[i][j] += t;
if(t == dp[i+1][j-1])
mark[i][j].pos = j-1;
if(t == dp[i+1][j])
mark[i][j].pos= j;
if(t == dp[i+1][j+1])
mark[i][j].pos = j+1;
}
}
}
int sx = 1;
int Max = dp[1][1];
for(int i = 2; i <= M; i++)
{
if(Max >= dp[1][i])//优先选择最右边的线
{
Max = dp[1][i];
sx = i;
}
}
printf("Case %d\n", k++);
printf("%d", sx);
int row = 1, cul = sx;//行号 列号
while(1)
{
if(row == N) break;
cul = mark[row][cul].pos;//下一列号
printf(" %d", cul);
row++;
}
printf("\n");
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &N, &M);
getMap();
solve();
}
return 0;
}
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