圆排列问题

#include "stdio.h"#include "math.h"#define MAX 10int n;  //圆的个数float min = 10000;  //当前圆排列的最小长度int r[MAX];  //各个圆的半径int bestr[MAX];   //最优排列下圆的半径float x[MAX];  //各个圆的圆心void swap(int &a, int &b){    int temp = a;    a = b;    b = temp;}//计算当前圆排列的长度void compute(){    int i;    float low = 0;  //圆位于原点右侧    float heigh = 0;    for(i=1; i<=n; i++)    {        if(x[i]-r[i]<low)  //寻找最左边的坐标，当圆位于原点左侧时才会更新            low = x[i]-r[i];        if(x[i]+r[i]>heigh)  //寻找最右边的坐标            heigh = x[i]+r[i];    }    if(heigh-low<min)  //如果当前圆排列的长度小于之前的最优值         min =  heigh - low;  //更新}//计算当前所选择圆的圆心float center(int i){    int j;    float temp = 0.0;    float valuex;    for(j=1; j<i; j++)    {        valuex = x[j] + 2.0*sqrt(r[j]*r[i]);  //寻找当前层下最右边的圆心        if(valuex>temp)            temp = valuex;    }    return temp;}void backtrack(int i){    if(i>n)  //如果到最后一个圆    {        compute();  //计算当前圆排列的最小长度        int  j;        for(j=1; j<=n; j++)              bestr[j] = r[j];    }    else    {          int j;        for(j=i; j<=n; j++)  //选择要排列的圆        {            swap(r[i], r[j]); //选择第j个圆            float centerx = center(i);  //计算圆心            if(centerx+r[i]<min)  //下界约束，当前圆的长度<最优值，才能产生最优值            {                x[i] = centerx;  //记录圆心                backtrack(i+1);  //搜索下一层            }            swap(r[i], r[j]);  //返回上一层        }    }}void circle(int n1, int r1[]){    n = n1;    int i;    for(i=1; i<=n; i++)        r[i] = r1[i];    backtrack(1);}int main(){    int n1 = 3;    int r1[]={0, 1, 1, 2};    circle(n1, r1);    printf("圆的个数为：%d\n", n);    printf("半径分别为：\n");    int i;    for(i=1; i<=n; i++)        printf("%d ", r1[i]);    printf("\n");    printf("最优排列下，圆的半径为：\n");    for(i=1; i<=n; i++)        printf("%d ", bestr[i]);    printf("\n");    printf("最小长度为：%.3f\n", min);    return 0;}