UVA 116 Unidirectional TSP

 C++代码

// [解题方法]
// 记忆化搜索（递归，子问题的结果用备忘录存起来，避免重复求解）
// 二维nxt数组按照题意记录路径
// dp[x][y](即dfs(x,y))表示从(x,y)走到最右边需要的最小花费

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define LL long long
#define inf 1e16

LL dp[11][111],w[11][111], mins, res[105];
int r, c;

struct point{
int x, y;
}nxt[11][111];

LL dfs (int x, int y)
{
if (y >= c) return (dp[x][y]=w[x][y]);
if (dp[x][y] < 1e16)
return dp[x][y];
int i;
LL tp = inf;
for (i = x-1; i <= x+1; i++)
{
int tr = i;
if (tr == 0) tr = r;
else if (tr > r) tr = 1;
LL tmp = dfs (tr, y+1) + w[x][y];
//更新路径(花费更小+字典序)
if (tmp < tp || (tmp == tp && tr < nxt[x][y].x)) {
nxt[x][y].x = tr;
nxt[x][y].y = y+1;
tp = tmp;
}
}
return (dp[x][y]=tp);
}

int main()
{
int i, j;
while (cin >> r >> c)
{
for (i = 1; i <= r; i++)
for (j = 1; j <= c; j++)
cin >> w[i][j], dp[i][j] = inf;
mins = inf;
int v;
memset (nxt, -1, sizeof(nxt));
for (i = 1; i <= r; i++)
{
LL tp = dfs (i, 1);
if (tp < mins) {
mins = tp;
v = i;
}
}
int tc = 1, k = 0;
while (v > 0)
{
res[k++] = v;
int tv = v;
v = nxt[tv][tc].x;
tc = nxt[tv][tc].y;
}
cout << res[0];
for (i = 1; i < k; i++) cout << ' ' << res[i];
cout << endl << mins << endl;
}
return 0;
}