UVA 818 Cutting Chains (DFS)

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What a find! Anna Locke has just bought several links of chain some of which may be connected. They are made from zorkium, a material that was frequently used to manufacture jewelry in the last century, but is not used for that purpose anymore. It has its very own shine, incomparable to gold or silver, and impossible to describe to anyone who has not seen it first hand.

Anna wants the pieces joined into a single end-to-end strand of chain. She takes the links to a jeweler who tells her that the cost of joining them depends on the number of chain links that must be opened and closed. In order to minimize the cost, she carefully calculates the minimum number of links that have to be opened to rejoin all the links into a single sequence. This turns out to be more difficult than she at first thought. You must solve this problem for her.

Input 

The input consists of descriptions of sets of chain links, one set per line. Each set is a list of integers delimited by one or more spaces. Every description starts with an integer n, which is the number of chain links in the set, where 1 ≤n ≤15. We will label the links 1, 2,..., n. The integers following n describe which links are connected to each other. Every connection is specified by a pair of integers i,j where 1 ≤i,j ≤n and i ≠j, indicating that chain links i and j are connected, i.e., one passes through the other. The description for each set is terminated by the pair -1 -1, which should not be processed.

The input is terminated by a description starting with n = 0. This description should not be processed and will not contain data for connected links.

Output 

For each set of chain links in the input, output a single line which reads

Set N: Minimum links to open is M

where N is the set number and M is the minimal number of links that have to be opened and closed such that all links can be joined into one single chain.

5 1 2 2 3 4 5 -1 -1
7 1 2 2 3 3 1 4 5 5 6 6 7 7 4 -1 -1
4 1 2 1 3 1 4 -1 -1
3 1 2 2 3 3 1 -1 -1
3 1 2 2 1 -1 -1
0

Set 1: Minimum links to open is 1
Set 2: Minimum links to open is 2
Set 3: Minimum links to open is 1
Set 4: Minimum links to open is 1
Set 5: Minimum links to open is 1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
int mp[15][15];
int vis[15];
int n,cnt;
bool ok(int s)//ok判断一个环接2个以上的环的不合法状态,
{
    REP(i,n)
    {
        if(s&(1<<i))  continue;
        int num=0;
        REP(j,n)
        {
            if(s&(1<<j))   continue;
            if(mp[i][j])  num++;
        }
        if(num>2)   return true;
    }
    return false;
}
bool dfs(int s,int now,int pre)
{
    vis[now]=1;
    REP(i,n)
    {
        if(!mp[now][i]||(s&(1<<i))||i==pre)   continue;
        if(vis[i])  return true;
        if(dfs(s,i,now))  return true;
    }
    return false;
}
bool circle(int s)//circle判断成环的不合法状态
{
    REP(i,n)
    {
        if(vis[i]||(s&(1<<i)))  continue;
        cnt++;
        if(dfs(s,i,-1))  return true;
    }
    return false;
}
int cal(int s)//计算open环的个数
{
    return s==0?0:cal(s>>1)+(s&1);
}
int solve()
{
    int ans=0x3f3f3f;
    int status=1<<n;
    REP(stau,status)//枚举状态1：代表open
    {
        cnt=0;CLEAR(vis,0);
        if(ok(stau)||circle(stau))  continue;//ok判断一个环接2个以上的环的不合法状态,circle判断成环的不合法状态
        if(cal(stau)>=cnt-1)
            ans=min(ans,cal(stau));
    }
    return ans;
}
int main()
{
    int x,y;
    int cas=1;
    while(~scanf("%d",&n)&&n)
    {
        CLEAR(mp,0);
        while(~scanf("%d%d",&x,&y)&&(x!=-1&&y!=-1))
            mp[x-1][y-1]=mp[y-1][x-1]=1;
        printf("Set %d: Minimum links to open is %d\n",cas++,solve());
    }
    return 0;
}

UVA 818 Cutting Chains (DFS)

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原文地址：http://blog.csdn.net/u013582254/article/details/40633575