UVa 116 - Unidirectional TSP
题目:求从左到右的一条路径上的加和最小。每次可以采cai取本行、上行和下行的走法。
分析:dp。数塔变形,由于要求最小序列,所以采取从右向左dp、可以保证右边的都是最小序列,每次记录后继节点,输出即可。
注意:在最上和最下两行有可能行的编号编程最大和最小的。
[cpp]
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int mat[ 12 ][ 102 ]={0};
int sum[ 12 ][ 102 ]={0};
int chi[ 12 ][ 102 ]={0};
void output( int r, int c, int m )
{
if ( c < m ) {
printf("%d ",r);
output( chi[r][c], c+1, m );
}else {
printf("%d\n",r);
}
}
int main()
{
int n,m;
while ( scanf("%d%d",&n,&m) != EOF ) {
memset( sum, 0, sizeof(sum) );
for ( int i = 1 ; i <= n ; ++ i )
for ( int j = 1 ; j <= m ; ++ j ) {
scanf("%d",&mat[i][j]);
}
for ( int i = m ; i >= 1 ; -- i )
for ( int j = 1 ; j <= n ; ++ j ) {
sum[j][i] = sum[j][i+1]+mat[j][i];
chi[j][i] = j;
int down = j+1;
if ( down > n ) down = 1;
if ( sum[j][i] == sum[down][i+1]+mat[j][i] ) {
sum[j][i] = sum[down][i+1]+mat[j][i];
if ( chi[j][i] > down )
chi[j][i] = down;
}
if ( sum[j][i] > sum[down][i+1]+mat[j][i] ) {
sum[j][i] = sum[down][i+1]+mat[j][i];
chi[j][i] = down;
}
int up = j-1;
if ( up < 1 ) up = n;
if ( sum[j][i] == sum[up][i+1]+mat[j][i] ) {
sum[j][i] = sum[up][i+1]+mat[j][i];
if ( chi[j][i] > up )
chi[j][i] = up;
}
if ( sum[j][i] > sum[up][i+1]+mat[j][i] ) {
sum[j][i] = sum[up][i+1]+mat[j][i];
chi[j][i] = up;
}
}
int minv = 0x7ffffff,space = 1;
for ( int i = 1 ; i <= n ; ++ i )
if ( minv > sum[i][1] ) {
minv = sum[i][1];
space = i;
}
output( space, 1, m );
printf("%d\n",minv);
}
return 0;
}
补充:软件开发 , C++ ,