地圖著色算法C語言實現[原創] 
       定理：任何平面地圖可以使用4種顏色給每個不同的城市著色，而保證相鄰的城市著不同的顏色。
       思路：把地圖上的每個城市抽象為一個點，并給每個城市編號，，相鄰的城市之間用直線連接。據此做出鄰接矩陣，若第i個城市與第j個城市相鄰，則metro[i][j]=1,否則metro[i][j]=0。
       算法：按照編號從小到大的順序檢查每個城市，對每個城市從1到4使用4種顏色著色，若當前顏色可用（即不與相鄰城市顏色相同），則著色；否則測試下一種顏色。
       程序：
＃i nclude <stdio.h>
#define N 21

int ok(int metro[N][N],int r_color[N],int current)
{/*測試當前著色方案是否可行*/
   int j;
   for(j=1;j<current;j++)
     if(metro[current][j]==1&&r_color[j]==r_color[current])
          return 0;/*城市相鄰且顏色相同*/
   return 1;
}

void go(int metro[N][N],int r_color[N],int sum,int current)
{
   int i;
   if(current<=sum)/*檢查所有城市*/
      for(i=1;i<=4;i++)/*測試每種顏色*/
      {
            r_color[current]=i;/*嘗試著色*/
            if(ok(metro,r_color,current))/*若嘗試成功*/
           {
                 go(metro,r_color,sum,current+1);/*檢查下一個城市*/
                 return;
          }
      }
}

void main()
{
   int r_color[N]={0};
   int i;
   int metro[N][N]={{0},/*鄰接矩陣*/
                            {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++)/*輸出著色方案*/
     printf("%3d",r_color[i]);
}

      中午吃飯時，想了一下，把上面程序中的go函數和ok函數合并一下，并且不使用遞歸，于是寫下了下面color函數的代碼，main函數不變，但調用方式變為color(metro,r_color,20);請朋友多批評

void color(int metro[N][N],int r_color[N],int sum)
{
   int i,j,k;
   for(i=1;i<=sum;i++)/*檢查所有城市*/
     for(j=1;j<=4;j++)/*對每個城市嘗試4種顏色的著色方案*/
     {
        r_color[i]=j;/*嘗試著色*/
        for(k=1;k<i;k++)/*檢查是否與相鄰城市顏色相同*/
            if(metro[i][k]==1&&r_color[k]==r_color[i])
               break;/*相同則跳出，此時有k<i，則下面條件不成立，繼續嘗試下一種顏色*/
        if(k>=i)/*若不相同，則使用當前顏色，并檢查下一個城市*/
            break;
     }
}