UVa 116 - Unidirectional TSP
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题目:给你一个n*m的数字表格,找到一条从左到右的路径,使得上面的数字和最小。
(每次可以从(i,j),走到(i,j+1),(i+1,j),(i-1,j)可以越界。)
分析:dp,动态规划。因为要字典序最小,所以采用从右向左的方式dp;
状态:f(i,j)表示走到(i,j)的最小和,则有转移方程:
f(i,j)= min(f(i+1,j+1),f(i,j+1),f(i-1,j+1));
记录路径输出即可。
说明:逆向dp保证字典序最小(后继最小),正向能保证每点前驱最小。
#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;int maps[11][101];int smap[11][101];int fath[11][101];int main(){int n,m;while (~scanf("%d%d",&n,&m)) {for (int i = 1 ; i <= n ; ++ i)for (int j = 1 ; j <= m ; ++ j)scanf("%d",&maps[i][j]);memset(smap, 0, sizeof(smap));for (int i = m ; i >= 1 ; -- i)for (int j = 1 ; j <= n ; ++ j) {smap[j][i] = smap[j][i+1]+maps[j][i];fath[j][i] = j;if (j > 1 && smap[j][i] >= smap[j-1][i+1]+maps[j][i]) {smap[j][i] = smap[j-1][i+1]+maps[j][i];fath[j][i] = j-1;}if (j == n && smap[j][i] >= smap[1][i+1]+maps[j][i]) {smap[j][i] = smap[1][i+1]+maps[j][i];fath[j][i] = 1;}if (j < n && smap[j][i] > smap[j+1][i+1]+maps[j][i]) {smap[j][i] = smap[j+1][i+1]+maps[j][i];fath[j][i] = j+1;}if (j == 1 && smap[j][i] > smap[n][i+1]+maps[j][i]) {smap[j][i] = smap[n][i+1]+maps[j][i];fath[j][i] = n;}}int spa = 1;for (int i = 2 ; i <= n ; ++ i)if (smap[spa][1] > smap[i][1])spa = i;int min = smap[spa][1];for (int i = 1 ; i <= m ; ++ i) {if (i < m) printf("%d ",spa);else printf("%d\n%d\n",spa,min);spa = fath[spa][i];}}return 0;}
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