Bicoloring-广搜

Bicoloring

In 1976 the ``Four Color Map Theorem" was proven with the assistance of a computer. This theorem states that every map can be colored using only four colors, in such a way that no region is colored using the same color as a neighbor region.

Here you are asked to solve a simpler similar problem. You have to decide whether a given arbitrary connected graph can be bicolored. That is, if one can assign colors (from a palette of two) to the nodes in such a way that no two adjacent nodes have the same color. To simplify the problem you can assume:

• no node will have an edge to itself.
• the graph is nondirected. That is, if a node a is said to be connected to a node b, then you must assume that b is connected to a.
• the graph will be strongly connected. That is, there will be at least one path from any node to any other node.
• 没有一个结点是和自己相连的
• 图是无向图,如果结点a和结点b相连,那么b就和a相连
• 至少有一条路径是从一个结点到其它任意一个结点

Input 

The input consists of several test cases. Each test case starts with a line containing the number n ( 1 < n < 200) of different nodes. The second line contains the number of edges l. After this, l lines will follow, each containing two numbers that specify an edge between the two nodes that they represent. A node in the graph will be labeled using a number a ( ).

An input with n = 0 will mark the end of the input and is not to be processed.

Output 

You have to decide whether the input graph can be bicolored or not, and print it as shown below.

Sample Input 

3
3
0 1
1 2
2 0
9
8
0 1
0 2
0 3
0 4
0 5
0 6
0 7
0 8
0

Sample Output 

NOT BICOLORABLE.
BICOLORABLE.

题意:给出n个结点,l条关系(ab结点之间有相连)

相连的两个结点之间颜色必须不同,问是否可以满足条件

那么如何判断是否可以满足呢?

用标记数组flag[]记录,第一个结点标记为1,即flag[0]=1,若0与1相连,并且结点1没有访问过,那么就将flag[1]标记为2,若1与3相连,且结点3没有访问过,那么flag[3]=1......若1与3相连,可是结点3已经访问过了,即flag[3]!=0,此时就判断f'lag[3]是否与flag[1]相等,若相等,则NOT BICOLORABLE.否则就继续查找与1相连的结点,如果后面没有与1相连的结点,那么这一层的dfs就跳出,继续查找与0相连的结点.

广搜代码(参考别人的),通过广搜知道了如何判断后,下面深搜代码自己写的

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int mapp[205][205],n;//mapp数组用来标记两节点相不相连
int flag[205];//flag数组用来标记结点是否走过
int bfs()
{
    int temp,i;
    queue<int>Q;
    Q.push(0);
    flag[0]=1;
    while(Q.empty()==0)
    {
        temp=Q.front();
        Q.pop();
        for(i=0;i<n;i++)
        {
            if(mapp[temp][i]==1)
            {
                if(flag[i]==0)
                {
                    Q.push(i);
                    if(flag[temp]==1)
                        flag[i]=2;
                    else if(flag[temp]==2)
                        flag[i]=1;
                }
                else if(flag[i]!=0)
                {
                    if(flag[i]==flag[temp])
                        return 1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    int l,i,x,y,ans;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        memset(flag,0,sizeof(flag));
        memset(mapp,0,sizeof(mapp));
        scanf("%d",&l);
        for(i=0;i<l;i++)
        {
            scanf("%d%d",&x,&y);
            mapp[x][y]=mapp[y][x]=1;
        }
        ans=bfs();
        if(ans==1)
        {
            printf("NOT BICOLORABLE.\n");
        }
        else
            printf("BICOLORABLE.\n");
    }
}

深搜代码

#include<stdio.h>
#include<string.h>
int l,mapp[205][205],flag[205],f,n;
void dfs(int v,int x)
{
    int i;
    flag[v]=x;
    for(i=0;i<n;i++)
    {
        if(mapp[v][i]==1||mapp[i][v]==1)
        {
            if(flag[i]==0)
            {
                if(flag[v]==1)
                    dfs(i,2);
                else if(flag[v]==2)
                    dfs(i,1);
            }
            else if(flag[i]!=0)
            {
                if(flag[v]==flag[i])
                {
                   f=1;
                   break;
                }

            }
        }
    }
    if(f==1)
      return;
}
int main()
{
    int x,y,ans,i;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        f=0;
        memset(mapp,0,sizeof(mapp));
        memset(flag,0,sizeof(flag));
        scanf("%d",&l);
        for(i=0;i<l;i++)
        {
            scanf("%d%d",&x,&y);
            mapp[x][y]=mapp[y][x]=1;
        }
            dfs(0,1);
            if(f==1)
                printf("NOT BICOLORABLE.\n");
            else
                printf("BICOLORABLE.\n");
    }
}