pku 1129 channel allocation 回溯解题报告
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pku 1129 channel allocation 回溯解题报告
典型的着色问题,用回溯解决
AC代码:
#include <stdio.h>
#include <string.h>
#define M 27
#define inf 10000000
int n, answer, x[M], map[M][M];
void generate_color(int i, int k)
{
int j;
while (1)
{
x[i] = (x[i] + 1) % (k + 1); //为x[i]选择颜色
if (x[i] == 0) //在k种颜色都不能匹配,证明只有k种方案不行
{
return;
}
for (j = 1; j <= n; j++) //匹配与i相连的结点,颜色相同,证明与x[i]相冲突
{
if (map[i][j] && x[i] == x[j])
{
break;
}
}
if (j == n + 1) //x[i]符合条件返回
{
return;
}
}
}
int coloring(int k)
{
int i, j;
for (j = 1; j <= n; j++)
{
x[j] = 0;
}
for (i = 1; i <= n; i++)
{
generate_color(i, k);
if (x[i] == 0)
{
return 0;
}
}
return 1;
}
int main()
{
//freopen("1.txt", "r", stdin);
int i, j, k;
while (scanf("%d", &n) != EOF && n != 0)
{
memset(map, 0, sizeof(map));
char c[M];
for (i = 1; i <= n; i++)
{
//利用邻接表来保存数据
scanf("%s", c);
k = strlen(c);
for (j = 2; j < k; j++)
{
map[i][c[j] - 'A' + 1] = 1;
}
}
for (i = n; i >= 1; i--)
{
if (coloring(i))
{
answer = i;
}
}
if (answer == 1)
{
printf("1 channel needed./n");
}
else
{
printf("%d channels needed./n", answer);
}
}
return 0;
}
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