趣味算法-城市之间最短总距离

城市之间最短总距离：图为联通图。

(1) 图中所有顶点集合为V, 最小生成树顶点集合为U初始时V包含所有顶点，U为空。

(2) 从V中选取一个顶点V0，将其加入U。

(3) 从V0的邻接顶点中选取边权值最小的Vn，得到最小生成树的一条边，将Vn加入集合U。

(4) 再从V-U中再选取一个与V0, Vn邻接的顶点，找出权值最小的边。

(5) 重复上述步骤。

#include <stdio.h>
#define MAX_DIST 65535

int calc(int arrGraph[5][5], int nPoints)
{
int i = 0;
int j = 0;
int nDist   = MAX_DIST;
int nPoint  = 0;
int nResult = 0;
int nLen = 0;

// there are two sets for U and V, we will select point from U and set to V
int *arrPointsU = (int*)malloc(nPoints*sizeof(int));
int *arrPointsV = (int*)malloc(nPoints*sizeof(int));

for (i = 0; i < nPoints; i++)
{
arrPointsU[i] = i;
arrPointsV[i] = -1;
}

// start from point 0
arrPointsV[0] = 0;
arrPointsU[0] = -1;
int nLenV = 1;

while (nLenV < nPoints)
{
nDist = MAX_DIST;
i = 0;
while(i < nLenV)
{
for (j = 0; j < nPoints; j++)
{
if (arrPointsU[j] == -1)
continue;
nLen = arrGraph[arrPointsV[i]][arrPointsU[j]];
if ((nLen > 0) && (nLen < nDist))
{
nPoint = j;
nDist = nLen;
}
}
i++;
}

if (nDist != MAX_DIST)
{
nLenV++;
arrPointsV[i] = nPoint;
arrPointsU[nPoint] = -1;
nResult += nDist;
printf("Point=%d\n", nPoint);
}
}
printf("Distance = %d \n", nResult);
free(arrPointsV);
free(arrPointsU);
return nResult;
}

int main()
{
int nDist = 0;
int i =0;
int arrTest[5][5]=
{
0, 2, 5, 0, 3,
2, 0, 0, 4, 0,
5, 0, 0, 0, 5,
0, 4, 0, 0, 2,
3, 0, 5, 2, 0,
};

nDist = calc(arrTest, 5);

printf("shortest distance=%d\n", nDist);

scanf("%d", &i);
return 0;
}

时间复杂度分析：与顶点的个数有关

n * ((1+2+ ... +n-1)*n) = n^2 * (n * (n-1)) / 2

空间复杂度 2*n 两个n个元素的数组保存集合U和V的数据。

可以用增大空间复杂的方式降低 时间复杂度，比如用二维数组保存边的权值，在循环时可降低时间复杂的