摘要： /*循环日程表问题(递归分治)*/#include #include#include#include#include#include#include#includeusing namespace std;#define maxn 2600//int f=0;int a[maxn][maxn];void cir(int i,int j,int n){ int p,q; if(n==2)//递归出口 { a[i][n]=j; a[i][n-1]=i; a[j][n]=i; a[j][n-1]=j; // pri... 阅读全文