uva216-网络连接

此题为暴力求解法回溯法的训练参考

翻译请戳 http://luckycat.kshs.kh.edu.tw/

解题思路

回溯。

可以用数组存储中间路径，

如果链接完毕且小于最小值，将中间路径存入最终路径。

详见代码。

代码

#include<stdio.h>
#include<math.h>
#include<string.h>
const int maxLen = 10;
typedef struct point {
    int x, y;
} Point;
Point setP[maxLen];
Point Path[maxLen];　　//存储“中间”路径
Point Final[maxLen];  //存储“最终”路径
bool connet[maxLen];
int n;
double min;
double dist(Point A, Point B)
{
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}
void Search(int start, int left, double res)
{
    if(res>min) return ;
    else if(left == 0) {
        min = res;
        for(int i=0; i<n; i++) Final[i] = Path[i];
        return ;
    }
    for(int i=0; i<n; i++) if(!connet[i]) {
        double mid = res;
        mid += (dist(setP[start], setP[i]) + 16.0);
        connet[i] = true;
        Path[n-left] = setP[i];
        Search(i, left-1, mid);
        connet[i] = false;
    }
}
int main()
{
    scanf("%d", &n);
    int tot = 0;
    while(n != 0) {
        memset(connet, 0, sizeof(connet));
        min = 1e9;
        for(int i=0; i<n; i++) {
            int x, y;
            scanf("%d%d", &x, &y);
            Point temp;
            temp.x=x; temp.y=y;
            setP[i] = temp;
        }
        for(int i=0; i<n; i++) {
            Path[0] = setP[i];
            connet[i] = true;
            Search(i, n-1, 0);
            connet[i] = false;
        }
        tot++;
        printf("**********************************************************\n");
        printf("Network #%d\n", tot);
        for(int i=n-1; i>0; i--) {
            printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",
              Final[i].x, Final[i].y, Final[i-1].x, Final[i-1].y, dist(Final[i], Final[i-1])+16.0);
        }
        printf("Number of feet of cable required is %.2lf.\n", min);
        scanf("%d", &n);
    }
    return 0;
}