sunfruit — Mon, 23 Oct 2006 13:17:00 GMT

--sunfruit

求上�?点到其他各点的最短�\径，依据图论知识建立矩阵模型�Q�进一步得��C��码如�?br />
public class ShortPathA {

private static int[][]
      a = {
      {0, 50, 10, 100000, 45, 100000}, {100000, 0, 15, 100000, 10, 100000}, {20, 100000, 0, 15, 100000, 100000}, {
      100000, 20, 100000, 0, 35, 100000}, {100000, 100000, 1000000, 30, 0, 100000}, {100000, 100000, 100000, 3, 100000, 0}
};

private static boolean[] mark = new boolean[a.length];
public ShortPathA() {
    int Vo = 0; //源点
    //源点到其他各点的距离
    int[] b = new int[a.length];
    DynArrayInt S = new DynArrayInt();
    for (int i = 0; i < a.length; i++) {
      mark[i] = false;
      //b[i] = a[Vo][i];
    }
    int best = -1;
    mark[0] = true;
    b[0] = 0; //{0为源点}
    while (best != 0) {
      best = 0;
      int best_j = 0;
      for (int i = 0; i < b.length; i++)
      {
        if (mark[i]) //{�Ҏ��一个已计算出最短�\径的点}
        {
          for (int j = 0; j < b.length; j++) {
            if ( (!mark[j]) && (a[i][j] > 0)) {
              if ( (best == 0) || (b[i] + a[i][j] < best)) {
                best = b[i] + a[i][j];
                best_j = j;
              }
            }
          }
        }
      }
      if (best > 0) {
        b[best_j] = best;
        mark[best_j] = true;
      }

}
System.out.println(java.util.Arrays.toString(b));
}

public static void main(String[] args) {
ShortPathA shortpath = new ShortPathA();
}

}

sunfruit 2006-10-23 21:17 发表评论

[原创]图论应用--一�W�画问题

sunfruit — Thu, 31 Aug 2006 06:20:00 GMT

--sunfruit

上图求一�W�画的�\径，利用图论的相关知识可以得到程序如下：

public class OnePath {

    private static int[][]
            links = { {0,1,1,0,0,0,1,0}, {1,0,0,1,0,0,0,1}, {1,0,0,1,1,1,0,0},
            {0,1,1,0,1,1,0,0}, {0,0,1,1,0,1,1,0}, {0,0,1,1,1,0,0,1}, {1,0,0,0,1,0,0,0}, {0,1,0,0,0,1,0,0}
    };

    public OnePath() {
        int sum = 0;
        //存放每个点的�?br />        int[] point = new int[links[0].length];
        for (int i = 0; i < links[0].length; i++) {
            int[] templink = links[i];
            for (int j = 0; j < links[0].length; j++) {
                point[i] += templink[j];
            }
            sum += point[i];
        }

        //计算度数是奇数点的个�?如果大于2则不能一�W�画
        int odt = 0;
        int start = -1;
        for (int i = 0; i < point.length; i++) {
            int mod = point[i] % 2;
            if (mod > 0) {
                //if(start==-1)
                    start = i;
                odt++;
            }
        }
        if(odt>2)
        {
          System.out.println("该图不能一�W�画");
          return;
        }
        int r = 0;
        //从一个奇数点开始计��?br />        int nowd=start;
        System.out.print(nowd+1);
        while (sum > 0) {
            r=0;
            //对于��L��nowd ��查当前的点r 是否合�?br />            //links[nowd][r]==0 判断是否有可以走的没有用�q�的�U��\
            //(point[r]<=1 && sum!=2) 判断是否是最后一�ơ，如果不是最后一�ơ，那么避开度数�?的点
            while (links[nowd][r]==0 || (point[r]<=1 && sum!=2)) {
                r++;
            }
            links[nowd][r]=0; //已经用过的线�?br />            links[r][nowd]=0; //已经用过的线�?links[nowd][r] links[r][nowd]互�ؓ往�q��\�U?用过1->2那么2->1也作废了
            sum=sum-2; //��d��数减2 因�ؓ�?->2 消耗了1的度�?的度
            point[nowd]--; //��L��和终点的度都�? 1->2 那么1的度�?的度都减1
            point[r]--; //��L��和终点的度都�? 1->2 那么1的度�?的度都减1
            nowd =r; //讄��新的��L��
            System.out.print("->"+(r+1));
        }
    }

    public static void main(String[] args) {
        new OnePath();
    }

}

sunfruit 2006-08-31 14:20 发表评论

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[原创]图论应用--一�W�画问题