10004 - 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.

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 (0 ≤ a < n).

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

0 1

1 2

2 0

0 1

1 2

0 1

0 2

0 3

0 4

0 5

0 6

0 7

0 8

Sample Output

NOT BICOLORABLE.

BICOLORABLE.

题目大意：用两种颜色给点涂色，任意相邻两点的颜色不能相同。

#include<stdio.h>
#include<memory.h>int edge[201][201];
int node[201];
int flag; //0 没相同的  1 表示有相同的 void DFS(int n,int begin,int color){       //n 点数  begin 当前遍历点  color 上一个点的涂抹颜色 1 2 color=(node[begin]==2?1:2);         int i;for(i=0;i<n;i++){if(edge[begin][i]){if(!node[i]){node[i]=color;DFS(n,i,color);}else if(node[i]==node[begin]){flag = 1;return ;}}}
}int main(){int n;  //节点数 while( ~scanf("%d",&n) && n ){int i,l; scanf("%d",&l);        //边数memset(edge,0,sizeof(edge));memset(node,0,sizeof(node));int x,y;for(i=0; i<l ; i++){scanf("%d%d",&x,&y);edge[x][y]=1;edge[y][x]=1;}flag = 0;node[x] = 1;DFS(n,x,1);if(!flag)printf("BICOLORABLE.\n");elseprintf("NOT BICOLORABLE.\n");      }return 0;
}

10004 - Bicoloring相关推荐

UVA 10004 - Bicoloring
模拟染色,因为只有两种颜色,所以分别用 0. 1 代表这两种颜色,然后从0开始深搜,如果每个点都能染上色,且相邻两点的颜色不同,则符合要求. #include<stdio.h>#incl ...
UVA 10004 Bicoloring
UVA_10004 对于这个题目,我们可以直接模拟染色的过程即可,如果发现要涂某一块时这个块已经被涂了色,并且与我们要使用的颜色不同的话,就说明这个图不能被染成BICOLORABLE的. #inclu ...
UVA 10004 Bicoloring （二分染色）
题目链接:https://vjudge.net/problem/UVA-10004 题意:给你一个连通图,问能否将每个节点染成两种颜色之一,使得每条边两边的节点颜色不同. (本质就是二分图的判定,有个 ...
π-Algorithmist分类题目（2）
原题网站:Algorithmist,http://www.algorithmist.com/index.php/Main_Page π-Algorithmist分类题目(2) Set Theory U ...
（Step1-500题）UVaOJ+算法竞赛入门经典+挑战编程+USACO
下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成.打牢基础,厚积薄发. 一.UVaOJ http://uva.onlinej ...
UVa Online Judge 工具網站
UVa Online Judge 工具網站转自http://www.csie.ntnu.edu.tw/~u91029/uva.html Lucky貓的ACM園地,Lucky貓的 ACM 中譯題目 M ...
算法竞赛入门经典+挑战编程+USACO
下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成.打牢基础,厚积薄发. 一.UVaOJ http://uva.onlinej ...
sicily题目分类
sicily题目分类 1. 编程入门 2. 数据结构 3. 字符串 4. 排序 5. 图遍历 6. 图算法 7. 搜索:剪枝,启发式搜索 8. 动态规划/递推 9. 分治/递归 10. 贪心 11. ...
[sicily]部分题目分类
sicily题目分类 1. 编程入门 2. 数据结构 3. 字符串 4. 排序 5. 图遍历 6. 图算法 7. 搜索:剪枝,启发式搜索 8. 动态规划/递推 9. 分治/递归 10. 贪心 11. ...

10004 - Bicoloring

10004 - Bicoloring相关推荐

最新文章

热门文章