Ultraviolet A 539-the Settlers of Catan, simple backtracking

539- Settlers of Catan

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Question link:

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

Question type: backtracking

Original question:

WithinSettlers of Catan, The 1995 German game of the year, players attempt to dominate an island by building roads, settlements and cities within SS 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 vicw.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.

o        o -- o        o \      /      \      /  o -- o        o -- o /      \      /      \o        o -- o        o -- o               \      /                o -- o

Sample input:

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

Sample output:

212

Question:

Enter an undirected graph and then output the longest path of the graph. Point can be repeated, but the side cannot be repeated.

Ideas and summary:

A simple backtracking recursive question is to enumerate all vertices directly, then start recursive search from this point, record the number of paths, and finally update and maintain a maximum value.

Code:

#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 LOCAL    freopen("input.txt","r",stdin);#endif    int 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\n",maxNum);    }     return 0;}

-- The meaning of life is to give it meaning.

 

Original Http://blog.csdn.net/shuangde800 ,
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