TJU1167 The Settlers of Catan +比赛总结
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 Specification
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 Specification
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
o o--o o
\ / \ /
o--o o--o
/ \ / \
o o--o o--o
\ /
o--o
比赛是看了这题以为是图论就跳过去了,赛后请教了looker,暴搜0MS过的,仅仅40行代码啊。
伤心啊,我比赛的时候怎么就没想到呢。原因:还没培养好递归的思想,数据结构不太强,要加油啊。
顺便说一下这次比赛,E G很水,大家基本都过了。
A字符串处理,刚开始处理得很头大,交上去WA了,原来把小写的‘k’写成‘K’了,粗心!
B我用快排做的,是Special judge,没看题上的“Print a blank line after each test case, even after the last one.”贡献了一次WA
H并查集实现,先对每条路的负载从大到小排序,依次加入每条路径直到起点和终点能够连通,就输出当前路径的负载
C、D、F赛时没实现,上面是C题,D题Team Queue要用链表实现,最后一个小时都贡献到这题上了,结果没有攻克
F题还没读题,superbin 比赛时8题全AC了,创一时佳话,膜拜一下。据悉F题不难,不过题目挺长的啊,呵呵。
贴下The Settlers of Catan (题意一笔画图)代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int map[30][30],max,n;
bool visit[3000];
int make(int id,int i)
{
if (id > i) return id+i*100;
else return id*100+i;
}
void dfs(int id,int m)
{
if (m > max) max = m;
int i;
for (i = 0; i< n; i++){
if (id != i && map[id][i] == 1 && visit[make(id,i)] == false){
visit[make(id,i)] = true;
dfs(i,m+1);
visit[make(id,i)] = false;
}
}
}
int main()
{
int t,i,x,y;
while (scanf ("%d%d",&n,&t) && n+t)
{
memset(map,0,sizeof(map));
for (i = 0; i< t; i++){
scanf ("%d%d",&x,&y);
map[x][y] = map[y][x] = 1;
}
max = 0;
for (i = 0; i< n; i++)
dfs(i,0);
printf ("%d\n",max);
}
return 0;
}