电路板排列问题

#include "stdio.h"#include "string.h"#define MAX 10       int x[MAX];  //当前电路板排列int bestx[MAX];  //最优电路板排列int n;  //电路板个数int bestd = 10000;  //最优密度，电路板密度为跨越相邻电路板最多连线数int m;  //连接块数int total[MAX];  //total[j]表示连接块j中电路板的个数int now[MAX];   //now[j]表示当前解下，连接块j中电路板的个数int b[9][6];  //连接块数组void swap(int &a, int &b){    int temp = a;    a = b;    b =temp;}//搜索排列树//第i个电路板，当前排列密度为cdvoid backtrack(int i, int cd){    if(i==n)  //当到达叶结点    {        bestd = cd;  //算法仅完成比当前解更优的解，所以cd肯定优于bestd        memcpy(bestx, x, sizeof(x));  //拷贝最优解向量    }    else    {        int j;        int ld;        for(j=i; j<=n; j++) //下一步选哪个电路板        {            //计算增加了电路板x[j]后的连线密度            ld = 0;            int k;            for(k=1; k<=m; k++)            {                now[k] += b[x[j]][k];  //计算当前包含在连接块k中的电路板个数                if(now[k]>0 && now[k]!=total[k]) //满足此条件                    ld++;  //连线密度增加            }            if(cd>ld) //更新ld                ld = cd;            if(ld<bestd)  //如果当前连线密度小于最优值，才可能产生最优值，搜索子树            {                swap(x[i], x[j]);                      backtrack(i+1, ld);    //搜索下一个结点                swap(x[i], x[j]);            }            //恢复状态，为返回上一层做准备            for(k=1; k<=m; k++)            {                now[k] -= b[x[j]][k];            }        }    }}void arrangement(int b1[9][6], int n1, int m1){    n = n1;    m = m1;    int i, j;    for(i=0; i<=n; i++)        for(j=0; j<=m; j++)            b[i][j] = b1[i][j];    memset(now, 0, sizeof(now));    memset(total, 0, sizeof(total));    //初始化x为单位排列并计算total    for(i=1; i<=n; i++)    {        x[i] = i;        for(j=1; j<=m; j++)            total[j] += b[i][j];    }    backtrack(1, 0);}int main(){    int n1 = 8;    int m1 = 5;    int b1[9][6] = {        {0, 0, 0, 0, 0, 0},        {0, 0, 0, 1, 0, 0},        {0, 0, 1, 0, 0, 0},        {0, 0, 1, 1, 1, 0},        {0, 1, 0, 0, 0, 0},        {0, 1, 0, 0, 0, 0},        {0, 1, 0, 0, 1, 0},        {0, 0, 0, 0, 0, 1},        {0, 0, 0, 0, 0, 1}    };    arrangement(b1, n1, m1);    printf("电路板个数：%d\n", n);    printf("连接块个数：%d\n", m);    printf("连接块情况：\n");    printf("N1={4,5,6},N2={2,3},N3={1,3},N4={3,6},N5={7,8}\n");    int i, j;    for(i=1; i<=n; i++)    {        for(j=1; j<=m; j++)            printf("%d ", b[i][j]);        printf("\n");    }    printf("最优密度为：%d\n", bestd);    printf("最优排列为：\n");    for(i=1; i<=n; i++)        printf("%d ", bestx[i]);    printf("\n");    return 0;}