定理：任何平面地图可以使用4种颜色给每个不同的城市着色，而保证相邻的城市着不同的颜色。
思路：把地图上的每个城市抽象为一个点，并给每个城市编号，，相邻的城市之间用直线连接。据此做出邻接矩阵，若第i个城市与第j个城市相邻，则metro[i][j]=1,否则metro[i][j]=0。
算法：按照编号从小到大的顺序检查每个城市，对每个城市从1到4使用4种颜色着色，若当前颜色可用（即不与相邻城市颜色相同），则着色；否则测试下一种颜色。
程序：
＃i nclude <stdio.h>
#define N 21
int ok(int metro[N][N],int r_color[N],int current)
{/*测试当前着色方案是否可行*/
int j;
for(j=1;j<current;j++)
if(metro[current][j]==1&&r_color[j]==r_color[current])
return 0;/*城市相邻且颜色相同*/
return 1;
}
void go(int metro[N][N],int r_color[N],int sum,int current)
{
int i;
if(current<=sum)/*检查所有城市*/
for(i=1;i<=4;i++)/*测试每种颜色*/
{
r_color[current]=i;/*尝试着色*/
if(ok(metro,r_color,current))/*若尝试成功*/
{
go(metro,r_color,sum,current+1);/*检查下一个城市*/
return;
}
}
}
void main()
{
int r_color[N]={0};
int i;
int metro[N][N]={{0},/*邻接矩阵*/
{0,1,1,1,1,1,1},
{0,1,1,1,1},
{0,1,1,1,0,0,1},
{0,1,1,0,1,1},
{0,1,0,0,1,1,1,0,0,1,0,0,0,0,0,0,1},
{0,1,0,1,0,1,1,1,1,1},
{0,0,0,0,0,0,1,1,1},
{0,0,0,0,0,0,1,1,1,1,0,0,1},
{0,0,0,0,0,1,1,0,1,1,0,0,1,1,1,0,1},
{0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,1,0,0,0,1},
{0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,1},
{0,0,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0,0,0,1,1},
{0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1},
{0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,1},
{0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,0,1},
{0,0,0,0,1,0,0,0,1,0,0,0,0,1,1,1,1},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1},
{0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,1,0,0,1,1,1},
{0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1}};
go(metro,r_color,20,1);
printf("\n");
for(i=1;i<=20;i++)/*输出着色方案*/
printf("%3d",r_color[i]);
}
中午吃饭时，想了一下，把上面程序中的go函数和ok函数合并一下，并且不使用递归，于是写下了下面color函数的代码，main函数不变，但调用方式变为color(metro,r_color,20);请朋友多批评
void color(int metro[N][N],int r_color[N],int sum)
{
int i,j,k;
for(i=1;i<=sum;i++)/*检查所有城市*/
for(j=1;j<=4;j++)/*对每个城市尝试4种颜色的着色方案*/
{
r_color[i]=j;/*尝试着色*/
for(k=1;k<i;k++)/*检查是否与相邻城市颜色相同*/
if(metro[i][k]==1&&r_color[k]==r_color[i])
break;/*相同则跳出，此时有k<i，则下面条件不成立，继续尝试下一种颜色*/
if(k>=i)/*若不相同，则使用当前颜色，并检查下一个城市*/
break;
}
}
