UVA - 116 - Unidirectional TSP （简单DP + 打印路径）

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题目传送： UVA - 116

思路：可以定义状态为dp[i][j] 为从第i行第j列开始往后走到第n列（总共n列）的最小值（赋初始值为无穷），且状态方程很好推出来：dp[i][j] = a[i][j] + max(dp[i-1][j+1], dp[i][j+1], dp[i+1][j+1]); 最后最优解 ans = max(dp[i][1])(1<=i<=m);

不过这题难点不在这里，而是可能有多组最小值，输出字典序最小的那组；

这里要注意好递推的方向，只能从右往左递推列数，而如果从左往右递推则不能满足字典序

AC代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x7fffffffusing namespace std;LL dp[15][105];int a[15][105];int beh[15][105];int n, m;int main() {while(scanf("%d %d", &m, &n) != EOF) {for(int i = 1; i <= m; i++) {for(int j = 1; j <= n; j++) {scanf("%d", &a[i][j]);dp[i][j] = INF;}}memset(beh, 0, sizeof(beh));for(int i = 1; i <= m; i++) {dp[i][n] = a[i][n];}for(int j = n - 1; j >= 1; j--) {for(int i = 1; i <= m; i++) {for(int k = -1; k <= 1; k++) {int t = i + k;if(t == 0) t = m;else if(t == m + 1) t = 1;if(a[i][j] + dp[t][j+1] < dp[i][j]) {dp[i][j] = a[i][j] + dp[t][j+1];beh[i][j] = t;}else if(a[i][j] + dp[t][j+1] == dp[i][j] && t < beh[i][j]) {beh[i][j] = t;}}}}int ans = INF, ansi, ansj = 1;for(int i = 1; i <= m; i++) {if(dp[i][1] < ans) {ans = dp[i][1];ansi = i;}}while(1) {if(beh[ansi][ansj] == 0) {printf("%d\n", ansi);break;}printf("%d ", ansi);ansi = beh[ansi][ansj];ansj ++;}printf("%d\n", ans);}return 0;}

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