UVA - 10004 Bicoloring
2014-09-11 22:16
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Bicoloring |
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, llines 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: 二染色no node will have an edge to itself. 不存在环
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. 图是连通
the graph will be strongly connected. That is, there will be at least one path from any node to any other node. 双向环
如果没有染色,将它染色,而且颜色相反
如果已经染色,是否与现在染色的点的颜色相同,相同,则退出
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #define N 1000 using namespace std; int vis ; int color ; int a ; int n,m; int dfs(int u) { for (int i = 0; i < n; i++) { if (a[u][i]) { if (vis[i] == 0) { vis[i] = 1; color[i] = !color[u]; dfs(i); } else if (color[u] == color[i]) return 0; } } return 1; } int main() { while (scanf("%d",&n) != EOF) { if (n == 0) break; memset(vis,0,sizeof(vis)); memset(a,0,sizeof(a)); scanf("%d",&m); for (int i = 0; i < m; i++) { int u, v; scanf ("%d%d",&u,&v); a[u][v] = a[v][u] = 1; } color[0] = 1; vis[0] = 1; if (dfs(0)) printf("BICOLORABLE.\n"); else printf("NOT BICOLORABLE.\n"); } return 0; }
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