UVA 116 Unidirectional TSP 经典dp题
题意:找最短路,知道三种行走方式,给出图,求出一条从左边到右边的最短路,且字典序最小。
用dp记忆化搜索的思想来考虑是思路很清晰的,但是困难在如何求出字典序最小的路。
因为左边到右边的字典序最小就必须从左边开始找,于是我们可以换个思路,dp时从右边走到左边,这样寻找路径就可以从左向右了。
代码:
/* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: uva116.cpp * Create Date: 2013-09-20 20:56:07 * Descripton: dp, memorial */ #include <cstdio> #include <algorithm> using namespace std; const int MAXN = 102; int dis[MAXN][MAXN], map[MAXN][MAXN], n, m; int cg(int x) { if (x == 0) x = n; else if (x == n + 1) x = 1; return x; } int dp(int x, int y) { x = cg(x); if (dis[x][y] != -0xffffff) return dis[x][y]; return dis[x][y] = map[x][y] + min(min(dp(x - 1, y + 1), dp(x, y + 1)), dp(x + 1, y + 1)); } void print(int x, int y) { if (y < m) printf("%d ", x); else { printf("%d\n", x); return; } int a[3] = {cg(x - 1), cg(x), cg(x + 1)}; sort(a, a + 3); int tt = dis[x][y] - map[x][y]; if (tt == dis[a[0]][y + 1]) print(a[0], y + 1); else if (tt == dis[a[1]][y + 1]) print(a[1], y + 1); else print(a[2], y + 1); } int main() { while (scanf("%d%d", &n, &m) != EOF) { for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { scanf("%d", &map[i][j]); if (j == m) dis[i][j] = map[i][j]; else dis[i][j] = -0xffffff; } int Min = 0xffffff, t, rx, ry; for (int i = 1; i <= n; i++) { t = dp(i, 1); if (t < Min) rx = i, Min = t; } print(rx, 1); printf("%d\n", Min); } return 0; }