uva 193 Graph Coloring(DFS)
2013-08-04 13:42
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Graph Coloring |
The coloring is restricted by the rule that no two connected nodes may be black.
![](http://uva.onlinejudge.org/external/1/193img1.gif)
Figure: An optimal graph with three black nodes
Input and Output
The graph is given as a set of nodes denoted by numbers ![](http://uva.onlinejudge.org/external/1/193img2.gif)
,
![](http://uva.onlinejudge.org/external/1/193img3.gif)
,
and a set of undirected edges denoted by pairs of node numbers
![](http://uva.onlinejudge.org/external/1/193img4.gif)
,
![](http://uva.onlinejudge.org/external/1/193img5.gif)
.
The input file contains m graphs. The number m is given on the first line. The first line of each graph contains n and k, the number of nodes and the number of edges, respectively. The following k lines contain
the edges given by a pair of node numbers, which are separated by a space.
The output should consists of 2m lines, two lines for each graph found in the input file. The first line of should contain the maximum number of nodes that can be colored black in the graph. The second line should contain one possible optimal coloring.
It is given by the list of black nodes, separated by a blank.
Sample Input
1 6 8 1 2 1 3 2 4 2 5 3 4 3 6 4 6 5 6
Sample Output
3 1 4 5
题目大意:给出一些相邻的点,然后要给点染色,染成黑色,要求相邻的点不能同时染成黑色,问最多染几个点,并且输出字典序最大的。
解题思路:读入点之间的关系后,单纯的用DFS去搜索,直到收索到最后一个点。
#include <stdio.h>
#include <string.h>
#define N 105
int g
, vis
, rec
;
int n, n_rec, m;
int judge(int k, int sum){
for (int i = 0; i <= sum; i++)
if (g[k][vis[i]])
return 1;
return 0;
}
void DFS(int k, int cnt){
vis[cnt] = k;
for (int i = k + 1; i <= n; i++){
if (judge(i, cnt)) continue;
DFS(i, cnt + 1);
}
if (cnt >= n_rec){
n_rec = cnt;
memcpy(rec, vis, sizeof(vis));
}
}
int main(){
int t;
scanf("%d", &t);
while (t--){
// Init;
memset(g, 0, sizeof(g));
memset(rec, 0, sizeof(rec));
n_rec = 0;
// Read;
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i++){
int a, b;
scanf("%d%d", &a, &b);
g[a][b] = g[b][a] = 1;
}
for (int i = 1; i <= n; i++){
memset(vis, 0, sizeof(vis));
DFS(i, 0);
}
printf("%d\n", n_rec + 1);
for (int i = 0; i < n_rec; i++)
printf("%d ", rec[i]);
printf("%d\n", rec[n_rec]);
}
return 0;
}
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