UVa116 (单向TSP,多决策问题)

/*----UVa1347 单向TSP
用d(i,j)表示从格子(i,j)出发到最后一列的最小开销
则在(i,j)处有三种决策,d(i,j)转移到d(i-1,j+1),d(i,j+1),d(i+1,j+1),还需要一个数组来记录每一步决策过程
*/
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<vector>
#include<string.h>
#include<math.h>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 100+10;
int a[maxn][maxn], dp[maxn][maxn],c[maxn][maxn];
int n, m;
void print(int i, int j){
	if (j < m){
		printf("%d", i+1);
		printf(j == m - 1 ? "\n" : " ");
		print(c[i][j], j + 1);
	}
}
int main(){
	int i, j,k;
	while (~scanf("%d%d", &n, &m)){
		for (i = 0; i < n; i++)
		for (j = 0; j < m; j++)
			scanf("%d", &a[i][j]);
		for (j = m - 1; j >= 0; j--)
		for (i = 0; i < n; i++){
			if (j == m - 1) dp[i][j] = a[i][j]; //边界条件
			else{
				if (i == 0)k = n- 1;
				else k = i - 1;
				dp[i][j] = dp[k][j + 1] + a[i][j];
				c[i][j] = k;

				if (dp[i][j] == dp[i][j + 1] + a[i][j] && i<c[i][j])
					c[i][j] = i;
				if (dp[i][j]>dp[i][j + 1] + a[i][j]){
					dp[i][j] = dp[i][j + 1] + a[i][j];
					c[i][j] = i;
				}

				if (i == n - 1) k = 0;
				else k = i + 1;
				if (dp[i][j] == dp[k][j + 1] + a[i][j] && k<c[i][j])
					c[i][j] = k;
				if (dp[i][j]>dp[k][j + 1] + a[i][j]){
					dp[i][j] = dp[k][j + 1] + a[i][j];
					c[i][j] = k;
				}
			}
		}
		j = 0;
		for (i = 1; i < n; i++){
			if (dp[i][0]<dp[j][0])
				j = i;
		}
		print(j, 0);
		printf("%d\n", dp[j][0]);
	}
	return 0;
}