poj2353Ministry<DP>

链接:http://poj.org/problem?id=2353

题意:

给定一矩阵,求从第一层走到最后一层,的最小花费, 每次只能走到相邻的位置,,输出最小花费路径;

思路:

看完题想到 DP 方程为 dp[i][j] = min( dp[i][j-1], dp[i][j+1], dp[i-1][j] )+cost[i][j];

可是 dp[i][j] 与 dp[i][j-1]和dp[i][j+1]有关感觉不靠谱,可是又没想到别的有效的办法;

后来又看了下发现到达某一层后每次只会选择一个方向, 要么左要么右, 于是我们就可以做两次比较;

dp[i][j] = min( dp[i][j-1], dp[i-1][j] )+cost[i][j];

dp[i][j] = min( dp[i][j+1], dp[i-1][j] )+cost[i][j];

大家都称之为双向DP,姑且就这么叫吧;

坑爹的还有要记录路径,

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int dp[110][510];
int cost[110][510], way[110][510], n, m;
void Print( int r, int c )
{
    if( r == 0 ){ 
        printf("%d\n", c+1);    
        return;
    }
    int x = way[r][c]/m, y = way[r][c]%m;
    Print(x, y);
    printf("%d\n", c+1);
    
}
int main()
{
    while( scanf("%d%d", &n,&m) != EOF){
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)    
                scanf("%d", &cost[i][j] );
        memset( dp, 0x3f, sizeof dp );    
        for(int i = 0; i < m; i++)
            dp[0][i] = cost[0][i], way[0][i] = i;
        for(int i = 1; i < n; i++){
            dp[i][0] = dp[i-1][0]+cost[i][0];
            way[i][0] = (i-1)*m;    
            for(int j = 1; j < m; j++){
                int t1=dp[i-1][j]+cost[i][j];
                int t2=dp[i][j-1]+cost[i][j];
                if( t1<t2 )
                    dp[i][j] = t1, way[i][j] = (i-1)*m + j;    
                else
                    dp[i][j] = t2, way[i][j] = i*m + (j-1);        
            }
            for(int j = m-2; j >= 0; j-- ){
                if( dp[i][j] > (dp[i][j+1]+cost[i][j]) )
                    dp[i][j] = dp[i][j+1]+cost[i][j], way[i][j] = i*m + (j+1);
                
            }
        }
        int Min = 1<<30, w;
        for(int i = 0; i < m; i++){    
            if( dp[n-1][i] < Min )
                Min = dp[n-1][i], w = i;
        }
        Print( n-1, w );    
    }
    return 0;
}