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;

}