UVa 539：The Settlers of Catan, 简单回溯

UVa 539：The Settlers of Catan, 简单回溯2014-10-11 shuangde800 题目链接：

http://uva.onlinejudge.org/index.php？option=com_onlinejudge&Itemid=8&category=108&page=show_problem&problem=480

题目类型：回溯法

原题：

Within Settlers of Catan, the 1995 German game of the year, players attempt to dominate an island by building roads, settlements and cities across its uncharted wilderness.

You are employed by a software company that just has decided to develop a computer version of this game, and you are chosen to implement one of the game"s special rules:

When the game ends, the player who built the longest road gains two extra victory points.

The problem here is that the players usually build complex road networks and not just one linear path. Therefore, determining the longest road is not trivial （although human players usually see it immediately）.

Compared to the original game, we will solve a simplified problem here: You are given a set of nodes （cities） and a set of edges （road segments） of length 1 connecting the nodes. The longest road is defined as the longest path within the network that doesn"t use an edge twice. Nodes may be visited more than once, though.

Example: The following network contains a road of length 12.

oo -- oo
 //
o -- oo -- o
 //
oo -- oo -- o
 /
o -- o

样例输入：

3 20 11 215 160 21 22 33 43 54 65 76 87 87 98 109 1110 1211 1210 1312 140 0

样例输出：

题目大意：

输入一张无向图，然后要求输出这张图最长的一条路径。点可以重复走，但是边不能重复走。

URL：http://www.bianceng.cn/Programming/sjjg/201410/45705.htm

思路与总结：

很简单的一道回溯递归题，直接枚举所有点，然后从这个点开始递归搜索，记录路径数量，最后更新维护一个最大值。

代码：

#include<iostream>#include<cstdio>#include<cstring>#define MAXN 30using namespace std;int n,m, G[MAXN][MAXN], vis[MAXN][MAXN], maxNum;void dfs（int u, int num）{for（int v=0; v<n; ++v）{if（G[u][v] && ！vis[u][v]）{vis[u][v] = vis[v][u] = 1;dfs（v, num+1）;vis[u][v] = vis[v][u] = 0;}}if（num > maxNum） maxNum = num;}int main（）{#ifdef LOCALfreopen（"input.txt","r",stdin）;#endifint a,b;while（~scanf（"%d %d", &n, &m））{if（！n && ！m） break;memset（G, 0, sizeof（G））;for（int i=0; i<m; ++i）{scanf（"%d %d", &a, &b）;++G[a][b];++G[b][a];} maxNum = -2147483645;for（int i=0; i<n; ++i）{memset（vis, 0, sizeof（vis））;dfs（i, 0）;}printf（"%d
",maxNum）;} return 0;}