Sherlock Holmes, the well known detective, must solve a puzzling situation. He has n boxes B1, B2, .., Bn , where n is even, each containing m balls. The balls are white and black. Let Bi = (Wi, Bi) denote a box with Wi white balls and Bi black balls. He must partition the boxes into two sets, each consisting of n/2 boxes, such that either the white balls or the black balls hold amajority in both sets. If there is such a majority, let m1 and m2 denote the percentage of the majority balls in each set. Holmes must find very quickly the maximum possible value of min(m1, m2). Can you help Holmes? 

There are several testcases. Each data set in the file stands for a particular set of boxes. A data set starts with the number n (n < 10000) of boxes. Follows the number m (m < 10000) of balls, and for each box the number (< 10000) of white and black balls in this specific order. White spaces can occur freely in the input. The input data are correct and terminate with an end of file.

For each set of data the program prints the result to the standard output from the beginning of a line. The program prints the color (W or B) of the balls that hold the majority followed by the maximum value - if there is such a majority, or "No solution" (without quotes) if no majority can be obtained. An input/output sample is in the table below. There is a single data set that contains 4 boxes each containing 30 balls. The first box, for example, contains 17 white balls and 13 black balls. There is only one possible partition of the boxes (B1, B4), (B2, B3), the white balls holding majority. The result for the data set is the identifier W and the maximum value.

43017 1312 1820 1014 16

W 51.67

Southeastern Europe 2006

一开始给的中文题目 居然没有说平均分成两组。。。

只能rand()%2分组+卡时 Wa了3个点

后来平均分后 随机交换两组中的两个+卡次数

写得挺丑的。。。

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;struct self{int x,y,side;}s[10001];int m,n,a,b,c;double ans,z;int side=-1;int baizong,heizong,num;int l[2],r[2];int x,y;void tongji(){ if(l[0]>r[0]&&l[1]>r[1]) { double k=min((double)l[0]/(m/2*n),(double)l[1]/(m/2*n)); if(k>ans) { ans=k; side=0; } } if(l[0]
          
           
            ans) { ans=k; side=1; } }}void exchange(int x,int y){ l[0]-=s[x].x; r[0]-=s[x].y; l[1]+=s[x].x; r[1]+=s[x].y; s[x].side=1; l[1]-=s[y].x; r[1]-=s[y].y; l[0]+=s[y].x; r[0]+=s[y].y; s[y].side=0; }int main(){ srand(time(NULL)); while(scanf("%d",&m)==1) { scanf("%d",&n); num=0; ans=0; baizong=0; heizong=0; side=-1; l[0]=r[0]=l[1]=r[1]=0; for(a=1;a<=m;a++) { scanf("%d%d",&s[a].x,&s[a].y); baizong+=s[a].x; heizong+=s[a].y; } if(baizong>heizong)side=0; else side=1; for(a=1;a<=m/2;a++) { s[a].side=0; l[0]+=s[a].x; r[0]+=s[a].y; } for(a=m/2+1;a<=m;a++) { s[a].side=1; l[1]+=s[a].x; r[1]+=s[a].y; } while(num<=50000) { num++; x=rand()%m+1; while(s[x].side!=0)x=rand()%m+1; y=rand()%m+1; while(s[y].side!=1)y=rand()%m+1; z=ans; exchange(x,y); tongji(); if(ans>z)if(rand()%1000+1<=50)exchange(y,x);else; else if(rand()%1000+1>=50)exchange(y,x); } if(ans==0) cout<<"No solution"<<'\n'; else if(side==0) cout<<"W "; else cout<<"B "; if(ans!=0)cout<
            
             <
             
              <
              
               <<'\n'; } return 0;}