ZOJ1025 Wooden Sticks

典型的贪心算法，和前面1029一样也是工作时间调度问题。

#include  < iostream >
#include  < vector >
#include  < algorithm >
using   namespace  std;

const   int  MAXSIZE  =   5000 ;
struct  stick
{ // 木条
     int  length; // 长度
     int  width; // 宽度
     bool  isVisited;
}Sticks[MAXSIZE];

bool  lessThan( const  stick &  s1, const  stick &  s2)
{ // 按木条长度从大到小排序
     return  s1.length < s2.length;
}
int  main( void )
{
     int  cases,i,j,k,n;
    cin >> cases;
     for (i = 1 ;i <= cases; ++ i)
    {
        cin >> n;
        vector < stick >  stickVect;
         for (j = 0 ;j < n; ++ j)
        {
            cin >> Sticks[j].length >> Sticks[j].width;
            Sticks[j].isVisited  =   false ;
            stickVect.push_back(Sticks[j]);
        }
        sort(stickVect.begin(),stickVect.end(),lessThan); // 排序
         // 贪心法
         int  length,width,time = 0 ,count = 0 ;
         for  (j = 0 ;j < n; ++ j)
        {
             if  (stickVect[j].isVisited == false )
            { // 还未访问,作为新一轮的排头兵
                time ++ ; // 新一轮了，时间加
                length  =  stickVect[j].length;
                width  =  stickVect[j].width;
                stickVect[j].isVisited  =   true ;
                 for (k = 0 ;k < n; ++ k)
                {
                     if  (stickVect[k].isVisited == false )
                    {
                         if  (stickVect[k].length >= length && stickVect[k].width >= width)
                        { // 此木条可以和排头兵一起顺利通过
                            stickVect[k].isVisited  =   true ;
                            length  =  stickVect[k].length;
                            width  =  stickVect[k].width;
                        }
                    }
                }
            }
        }
        cout << time << endl;
    }
     return   0 ;
}