Unidirectional TSP UVA - 116

按照紫书的思路，从最终的状态回推，每次都是来源于三个方向，取出三个方向上最小的值和当前的值进行结合并且记录相应的编号。最后一起输出即可，具体实现见如下代码：

#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<deque>
using namespace std;

int m, n;
int data[15][110];
int ind[15][110];
int ans[15][110];

int main(){
	while (cin >> m >> n){
		memset(ind,-1,sizeof(ind));
		memset(ans, 0, sizeof(ans));
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				cin >> data[i][j];
			}
		}
		for (int j = n - 1; j >= 0; j--){//col
			for (int i = 0; i < m; i++){
				int row[3] = {i,i-1,i+1};
				if (row[1] < 0) row[1] = m-1;
				if (row[2] >= m) row[2] = 0;
				sort(row,row+3);
				ans[i][j] = 1 << 20;
				for (int k = 0; k < 3; k++){
					int temp = ans[row[k]][j + 1] + data[i][j];
					if (temp < ans[i][j]){
						ans[i][j] = temp;
						if (j!=n-1) ind[i][j] = row[k];
					}
				}
			}
		}
		int start;
		int temp = 1 << 20;
		for (int i = 0; i < m; i++){
			if (ans[i][0] < temp){
				temp = ans[i][0];
				start = i;
			}
		}
		cout << start+1;
		int j = 0;
		while (ind[start][j] != -1){
			cout << " " << ind[start][j]+1;
			start = ind[start][j];
			j++;
		}
		cout << endl;
		cout << temp << endl;
	}
	return 0;
}