uva 539 The Settlers of Catan

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

题目描述：

The Settlers of Catan

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.

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

Input 

The first line of each test case contains two integers: the number of nodes n ( ) and the number of edges m ( ). The next m lines describe the m edges. Each edge is given by the numbers of the two nodes connected by it. Nodes are numbered from 0 to n-1. Edges are undirected. Nodes have degrees of three or less. The network is not neccessarily connected.

Input will be terminated by two values of 0 for n and m.

Output 

Sample Input 

3 2
0 1
1 2
15 16
0 2
1 2
2 3
3 4
3 5
4 6
5 7
6 8
7 8
7 9
8 10
9 11
10 12
11 12
10 13
12 14
0 0

Sample Output 

2
12

题意：找存在的最长的欧拉回路。

题解：暴力深搜。

代码：

/*topcoder codeforce acmuva 10474 - Where is the Marble?no repeat edges between two vertexs*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<time.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;typedef struct Points
{int start;int end;bool used;Points(){start=0;end=0;used=false;}
}Points,*PointsLink;
Points List[200+5];bool Graph[25+5][25+5]={false};//true is connectedint n=0;//the count of the nodes,the node number begin with 0
int m=0;//the count of the edges
int MaxLen=0;/*DFS the graph and find the euler path*/
int DFS(int cur,int Len)
{if(Len>MaxLen){MaxLen=Len;}int i=0,j=0;for(i=0;i<=n-1;i++){if(Graph[cur][i])//connected{//not usedfor(j=0;j<=m-1;j++){if((cur==List[j].start&&i==List[j].end)||(cur==List[j].end&&i==List[j].start)){if(List[j].used==false){List[j].used=true;DFS(i,Len+1);List[j].used=false;break;//has found so it is no need to continue to find}}}}}return(0);
}/*for test*/
int test()
{return(0);
}/*main process*/
int MainProc()
{while(scanf("%d%d",&n,&m)!=EOF&&(n>0||m>0)){int i=0;int Start=0,End=0;memset(Graph,false,sizeof(Graph));MaxLen=0;for(i=0;i<=m-1;i++){scanf("%d%d",&Start,&End);Graph[Start][End]=true;Graph[End][Start]=true;List[i].start=Start;List[i].end=End;List[i].used=false;}if(m<=1){printf("%d\n",m);}else{int Len=0;for(i=0;i<=n-1;i++){Len=0;DFS(i,Len);}printf("%d\n",MaxLen);}}return(0);
}int main()
{MainProc();return(0);
}