UVa10004 Bicoloring(二分图判定)
2014-04-04 12:05
246 查看
UVa10004 Bicoloring(二分图判定)
分类: 图论2013-11-0613:18 182人阅读 评论(0) 收藏 举报
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 ( $0 \le a < n$).
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
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.
code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 200 + 10;
vector<int> G[maxn];
int n, m;
int color[maxn];
bool dfs(int v, int c)
{
color[v] = c; //把顶点v染成颜色c
for(int i=0; i< G[v].size(); ++i) {
//如果相邻的顶点同色,则返回false
if(color[G[v][i]] == c) return false;
//如果相邻的顶点还没有被染色,则染成-c
if(color[G[v][i]] == 0 && !dfs(G[v][i], -c)) return false;
}
//如果所有顶点都染过色了,则返回true
return true;
}
void solve()
{
for(int i=0; i<n; ++i) {
if(color[i] == 0) {
//如果顶点i还没被染色,则染成1
if(!dfs(i,1)) {
printf("NOT BICOLORABLE.\n");
return ;
}
}
}
printf("BICOLORABLE.\n");
}
int main()
{
int i, x, y;
while(scanf("%d",&n),n) {
scanf("%d",&m);
memset(color, 0, sizeof color );
for(i=0; i<=n; ++i) G[i].clear();
for(i=0; i<m; ++i) {
scanf("%d%d",&x,&y);
G[x].push_back(y);
G[y].push_back(x);
}
solve();
}
return 0;
}
相关文章推荐
- 二分图判定(UVA10004)(DFS或者BFS)
- UVa 10004 Bicoloring(二分图判定+DFS)
- POJ:2492-Bug's Life(二分图的判定)
- [hdu4598]二分图判定,差分约束
- HDU 2444 The Accomodation of Students二分图判定和匈牙利算法
- 点的双联通+二分图的判定(poj2942)
- hdu2444二分图判定 + 最大匹配
- hihocoder #1121 : 二分图一•二分图判定
- 二分图判定(染色问题)
- 099_二分图判定
- HDU 2444 The Accomodation of Students(二分图判定+最大匹配)
- [HihoCoder]#1121 : 二分图一•二分图判定
- UVa - 11396 Claw Decomposition 二分图的判定
- NP-Hard Problemd(二分图判定着色)
- poj2942[补图+点双连通分量+交叉染色法判定二分图(奇圈判定)]
- hihoCoder#1121(二分图判定)
- POJ 1112 Team Them Up! 二分图判定+01背包
- 二分图判定 (挑战程序设计竞赛)
- hdu2444 二分图判定+匹配
- hdu4751(二分图判定)