2018_1_23_poj_2258_回溯
2018-01-23 15:56
253 查看
The Settlers of Catan
Description
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.
Input
The input will contain one or more test cases.
The first line of each test case contains two integers: the number of nodes n (2<=n<=25) and the number of edges m (1<=m<=25). 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
For each test case, print the length of the longest road on a single line.
Sample Input
Sample Output
Source
Ulm Local 1998
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 1339 | Accepted: 872 |
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 input will contain one or more test cases.
The first line of each test case contains two integers: the number of nodes n (2<=n<=25) and the number of edges m (1<=m<=25). 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
For each test case, print the length of the longest road on a single line.
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
Source
Ulm Local 1998
#include<iostream> #include<string> #include<vector> #include<algorithm> #include<cstring> using namespace std; int n,m,best; int a[32][32]; bool read(){ scanf("%d%d",&n,&m); if(n==0&&m==0)return 0; fill(&a[0][0],&a[0][0]+32*32,0); int x,y; while(m--){ scanf("%d%d",&x,&y); a[x][y]=a[x][y]=1; } return 1; } void visit(int i,int l){ for(int j=0;j<n;j++) if(a[i][j]){ a[i][j]=a[j][i]=0; visit(j,l+1); a[i][j]=a[j][i]=1; } best=best<l?l:best; } void solve(){ best=0; for(int i=0;i<n;i++) visit(i,0); printf("%d\n",best); } int main(){ while(read())solve(); return 0; }
相关文章推荐
- 2018_1_23_poj_1011_sticks_枚举_递归_剪枝
- 2018_1_23_uva167_回溯打表_
- 2018_1_23_poj_2083
- 2018_1_24_poj_1040_回溯
- 2013_CSUST_3_23校内训练赛第一场【old】【hdu 3496、2191、4508、4506、2181 POJ 3264 3210 3094】
- POJ 2018 动态规划
- POJ 3009 Curling 2.0【带回溯DFS】
- 23 Out of 5 uva+回溯
- poj 2018 分类: poj 2015-04-22 13:48 32人阅读 评论(0) 收藏
- POJ 2488 DFS+回溯
- 2018_2_5_Anagrams by Stack_栈_模拟_回溯_递归
- Firetruck_dfs预处理_2018_2_23
- POJ 3694 Network tarjan求桥+回溯求祖先
- POJ 1321----棋盘问题(dfs+回溯)
- poj2018 Best Cow Fences
- poj 2018
- 2013_CSUST_3_23校内训练赛第一场【old】【hdu 3496、2191、4508、4506、2181 POJ 3264 3210 3094】
- poj 2488 dfs+回溯
- poj 3414 Pots【bfs+回溯路径 正向输出】
- POJ 2488 A Knight's Journey 递归回溯题解