Uva539 the Settlers of Catan

The Settlers of Catan

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

Inputthe input file will contain in one or more test cases.

The first line of each test case contains two integers: the number of nodesN() And the number of EdgesM(). The nextMLines describeMEdges. Each edge is given by the numbers of the two nodes connected by it. nodes are numbered from 0N-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 0NAndM.

Outputfor each test case, print the length of the longest road on a single line. 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 n nodes and an undirected graph of M edge (not necessarily connected) to find the longest path length. The edge cannot pass twice, but the vertex can pass through
// Limit: 2 <= n <= 25, 1 <= m <= 25, no self-ring or duplicate edge
// Algorithm: DFS

 

Update depth in DFS d

#include<cstdio>#include<cstring>#include<iostream>#include<string>#include<algorithm>using namespace std;const int maxn=26;int a[maxn][maxn];int n, m;int ans;void dfs(int i, int d){    for(int j=0; j<n; j++) if(a[i][j]) {        a[i][j]=0;        a[j][i]=0;        dfs(j, d+1);        a[i][j]=1;        a[j][i]=1;    }    ans=max(ans, d);}int solve(){    ans=0;    for(int i=0;i<n;i++)        dfs(i, 0);    return ans;}int main(){#ifndef ONLINE_JUDGE    freopen("./uva539.in", "r", stdin);#endif    while(scanf("%d%d", &n, &m)==2 && (n || m)) {        memset(a, 0, sizeof(a));        for(int i=0;i<m;i++) {            int x,y;            scanf("%d%d", &x, &y);            a[x][y]=1;            a[y][x]=1;        }        printf("%d\n", solve());    }    return 0;}

DFS returns the longest length.

#include<cstdio>#include<cstring>#include<iostream>#include<string>#include<algorithm>using namespace std;const int maxn=26;int G[maxn][maxn];int n, m;int dfs(int i){    int len=0;    for(int j=0; j<n; j++) if(G[i][j]) {        G[i][j]=G[j][i]=0;        len=max(len, dfs(j)+1);        G[i][j]=G[j][i]=1;    }    return len;}int solve(){    int ans=0;    for(int i=0;i<n;i++)        ans=max(ans, dfs(i));    return ans;}int main(){#ifndef ONLINE_JUDGE    freopen("./uva539.in", "r", stdin);#endif    while(scanf("%d%d", &n, &m)==2 && (n || m)) {        memset(G, 0, sizeof(G));        for(int i=0;i<m;i++) {            int x,y;            scanf("%d%d", &x, &y); G[x][y]=G[y][x]=1;        }        printf("%d\n", solve());    }    return 0;}