2013年湘潭邀请赛解题报告

              
A：
题目大意: Alice和Bob分别按照不同规则从一堆石子中取石子，问至少需要多少次能够把石子取玩。
#include<stdio.h>
#include <string.h>

int g[10000+100];
int f[10000+100];


int min(int x,int y)
{
 if(x<y) return x;
 return y;    
}

void  solve()
{
   g[1]=1;
   f[1]=1;
   
   int i,j;
   for(i=2;i<=10000;i++)
   {
      f[i]=i;
      for(j=1;j<=i;j=j*2)
      {
        f[i]=min(f[i],g[i-j]+1);                   
      }
      g[i]=i;
      for(j=1;j<=i;j=3*j) 
      {
          g[i]=min(g[i],f[i-j]+1); 
                          
      }
   }
}

int main()
{
    int t,n;
    
    solve();
    scanf("%d",&t);
    
    while(t--)
    {
       scanf("%d",&n); 
       
       printf("%d\n",f[n]);
               
    }
    
    
return 0;    
}

B：
题目大意：给定一棵带值二叉树，问至少需要修改多少节点的值能够使得二叉树成为二叉排序树。

解题思路：先使用中序遍历获得整个序列，然后求这个序列的最长上升子序列即可。
Ans = n – 最长上升子序列长度
时间复杂度：O(nlogn)

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 5010;
const int inf = 0x3f3f3f3f;
int g[N],d[N];

struct node{
   int key,lch,rch;       
}Q[N];

int a[N],pre[N],n,top;
int idx[N],m;

void dfs(int x)
{
  if(Q[x].lch!=0) dfs(Q[x].lch);   
    a[++top]=Q[x].key;
  
   if(Q[x].rch!=0) dfs(Q[x].rch);   
   // a[++top]=Q[x].key;
}

int lis()
{
    
     for(int i=1;i<=n;i++)
      g[i]=inf;
      
      int ans=0;
      
      for(int i=1;i<=n;i++)
      {
        int k=lower_bound(g+1,g+1+n,a[i])-g;
        d[i]=k;
        g[k]=a[i];
        ans=max(ans,d[i]);                 
                       
      }
      
  return n-ans;
    
}

int main()
{
    int t;
    scanf("%d",&t);
    
    while(t--)
    {
        scanf("%d",&n);
        
        memset(pre,0,sizeof(pre));
        
        for(int i=1;i<=n;i++)
        {
          scanf("%d%d%d",&Q[i].key,&Q[i].lch,&Q[i].rch);  
          pre[Q[i].lch]=i;
          pre[Q[i].rch]=i;
        }   
        
        int rt;
        
        for(int i = 1;i <= n; i++)
         if(pre[i]==0)
          {
           rt=i;
           break;             
          }
       top=0;
       dfs(rt); 
        
       // for(int i = 1; i <= n; i++) 
      //      printf("%d ", a[i] ); puts(""); 
        
        
      printf("%d\n",lis());
             
              
    }
    
    
return 0;    
}

D DNA
字典树+dp
package com.wy.tools;


import java.util.HashMap;
import java.util.Scanner;


public class Main {
   HashMap<Character,Integer> map=new HashMap<Character,Integer>();
   
   Node root;
   long[] f;
   private Node tmp;
   
   static Scanner s=new Scanner(System.in);
   
   public Main()
   {
	   String st;
	   
	   map.put('A', 0);
	   map.put('T', 1);
	   map.put('G', 2);
	   map.put('C', 3);
	   
	   int x;
	   int n;
	   
	   f=new long[110000];
	   root =new Node();
	   n=s.nextInt();
	   
	   while(n--!=0)
	   {
		   st=s.next();
		   StringBuffer sb=new StringBuffer(st);
		   sb.reverse();
		   x=s.nextInt();
		 //  insert(st,x);
		   insert(sb.toString(),x);
	   }
	   
	   st=s.next();
	   
	   int p;
	   
	   for(int i=0;i<st.length();i++)
	   {
		   tmp=root;
		   p=i;
		   
		   if(i!=0)
			   f[i]=f[i-1];
		   
		   while(p>=0)
		   {
			   tmp=tmp.son[map.get(st.charAt(p))];
			   
			   if(tmp==null)
				   break;
			   
			   if(tmp.val!=-1)
			   {
				  if(p==0)
					  f[i]=Math.max(f[i], tmp.val);
				  else
					  f[i]=Math.max(f[p-1]+tmp.val,f[i] );
			   }
			   p=p-1;
			 }
		  }
	   
	 //  for(int i=0;i<st.length();i++)
	//	     System.out.println(f[i]);

	   
	     System.out.println(f[st.length()-1]);
	     }
   
    public void insert(String s,int x)
    {
      	tmp=root;
      	for(int i=0;i<s.length();i++)
      	{
      		int idx=map.get(s.charAt(i));
      		if(tmp.son[idx]==null)
      		{
      			tmp.son[idx]=new Node();
      		}
      		tmp=tmp.son[idx];
      	}
      	tmp.val=Math.max(tmp.val, x);
    }
    
	public static void main(String[] args) {
		 int t=s.nextInt();
		 while(t--!=0)
		 {
			 new Main();
		 }
		
	
	}

}

class Node
{
	Node [] son=new Node[4];
	int val=-1;
}

/*
 * 
3
GG 100
ACG 50
GGAC 10000
GGGGACTACGG
GGCAGGTACGG
*/
E Edges of Triangle（未过）
#include <stdio.h>
#include <string.h>
#include <math.h>

struct Point
{
  double x,y;       
}point[5];


struct Line
{
   Point a,b;      
}line[5];


/*
Point intersection(Line u,Line v){  
    Point ret=u.a;  
    double t=((u.a.x-v.a.x)*(v.b.y-v.a.y)-(u.a.y-v.a.y)*(v.b.x-v.a.x))  
        /((u.a.x-u.b.x)*(v.b.y-v.a.y)-(u.a.y-u.b.y)*(v.b.x-v.a.x));  
    ret.x+=(u.b.x-u.a.x)*t;  
    ret.y+=(u.b.y-u.a.y)*t;  
    return ret;  
}  
*/

Point intersection(double a1,double b1,double c1,double a2,double b2,double c2){  
     
  Point ret;
  
  ret.x=(b2*c1-b1*c2)/(a1*b2-a2*b1);
  
  if(b1!=0)
    ret.y = (c1-a1*ret.x)/b1;
  else if(b2!=0)
    ret.y = (c2-a2*ret.x)/b2;
  
  return ret;
}


long long gcd(long long a,long long b)
{
  if(b==0) return a;
  return gcd(b,a%b);    
}

long long  Onside(int n)
{
    int i=0,ret=0;
    
    for(i=0;i<n;i++)
      ret+=gcd( (long long)100000000000*fabs(point[(i+1)%n].x-point[i%n].x) ,(long long)100000000000*fabs(point[(i+1)%n].y-point[i%n].y) );
    return ret/100000000000;
}



int main()
{
    int t,i;
    double a[5],b[5],c[5];
    
    scanf("%d",&t);
    
    while(t--)
    {
      //scanf("%lf%lf%lf%lf%lf%lf",&a1,&b1,&a2,&b2,&a3,&b3); 
      
      for(i=1;i<=3;i++)
      {
        scanf("%lf%lf%lf",&a[i],&b[i],&c[i]);
        
       // printf("(a)%lf (b)%lf (c)%lf\n",a,b,c);
       
      }  
      
       point[0]= intersection(a[1],b[1],c[1],a[2],b[2],c[2]);
       point[1]= intersection(a[2],b[2],c[2],a[3],b[3],c[3]);
       point[2]= intersection(a[1],b[1],c[1],a[3],b[3],c[3]);
    //  for(i=0;i<3;i++)
     //{
     //   printf("%lf %lf\n",point[i].x,point[i].y);
                        
   //  }
      
      printf("%lld\n",Onside(3));
      
      
          
              
    }  
    
    
 return 0;    
}
F:Five Tigers
模拟

H: Hurry up(三分)
#include <stdio.h>
#include <string.h>
#include <math.h>


double vr,vt;
double x3,y3;
double x4,y4;

double distance(double x1,double y1,double x2,double y2)
{
    return sqrt( (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1) );
}


double f(double x)
{
    double     t1= distance(x3,y3,x,0)/vr;
    double     t2= distance(x,0,x4,y4)/vt;
    return t1+t2;  
}

double min(double x,double y)
{
  if(x<y) return x;
  return y;
}

int main()
{
    int t;
	double t1;
    
    scanf("%d",&t);
    
    while(t--)
    {
       scanf("%lf%lf%lf%lf%lf%lf",&x3,&y3,&x4,&y4,&vr,&vt);   
       
       t1= distance(x3,y3,x4,y4)/vr;
       
       
       double   l=x3,r=x4;
       
       
       for(int i=0;i<100;i++)
       {
           double m1=l + (r-l)/3;
           double m2=r - (r-l)/3;  
           
           if(f(m1)<f(m2))  r=m2;
           else  l=m1;
       }
       
       printf("%.2lf\n",min(t1,f(l)));
       
       
           
              
    }
    
    
return 0;    
}
I:I Love Military Chess(模拟)
J:Jack’s Sequence
类型：构造
题目大意：给定一个合法的括号序列，求下一个字典序排列最小的合法括号序列。
解题思路：找出最后一个符合(xx(xx)xx)xx模型的子串并将其改为(xx)(((..)))即可，xx表示任意合法的括号序列。
(())()
()(())
可参考下面的博客
http://www.cnblogs.com/yefeng1627/archive/2013/05/12/3074443.html