uva 216 – getting in line

Contents

Problem

中文網址

Solution

建好邊後,DFS 找最佳解。

Code

UVa 216
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#include<cmath>

struct
{
int x, y;
double getDis(Point& p)
{
return sqrt((x - p.x)*(x - p.x) + (y - p.y)*(y - p.y));
}

}point[8];
double dis[8][8],min;
int n, seq[8], min_seq[8];
bool visit[8];
void dfs(int prev, int count, double len)
{
if (count == n)
{
if (min > len)
{
min = len;
for (int i = 0; i < n; i++)
min_seq[i] = seq[i];
}

return;
}

for (int i = 0; i < n; i++)
{
if (!visit[i])
{
visit[i] = true;
seq[count] = i;
dfs(i, count + 1, len + dis[prev][i]);
visit[i] = false;
}
}
}
int main()
{
int i, Case = 1;
while (scanf("%d", &n) && n)
{
for (i = 0; i < n; i++)
scanf("%d%d", &point[i].x, &point[i].y);

for (i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
dis[i][j] = dis[j][i] = point[i].getDis(point[j]);

min = 1e9;
for (i = 0; i < n; i++)
{
seq[0] = i;
visit[i] = true;
dfs(i, 1, 0);
visit[i] = false;
}

puts("**********************************************************");
printf("Network #%dn", Case++);
for (i = 0; i < n - 1; i++)
printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.n", point[min_seq[i]].x, point[min_seq[i]].y, point[min_seq[i + 1]].x, point[min_seq[i + 1]].y, dis[min_seq[i]][min_seq[i + 1]]+16);
printf("Number of feet of cable required is %.2lf.n", min+16*(n-1));
}

return 0;
}