20,190,810 summer camped exam Day2

Title \ (100 + 50 + 0 \)

LG1988 torch

Correct

1. (When problems do, prompted by the apparent \ (n \ times m \) in the range of 64-bit integer,) enumeration \ (n \ times m \)
   
2. Just enumeration product contains only 0 or 1, similar to binary, queue creating

Thinking

1. Reverse entitled seeking \ (m \) , but enumeration \ (n \ times m \)

\ (DHOJ \ 1403 \) or \ (JZOJ \ 3053 \)

Correct

1. 50% Manhattan distance, axes partition
   
2. Paint 100%, in some cases ( Interpretations ) +2 Manhattan distance, axes partition

Thinking

1. At first I thought a tree-DP, tune fast one hour discovery algorithms wrong
2. 50% think of Manhattan, but did not expect the addition principle of divide and conquer +
   
3. Paint not completely, 100 percent of the ignorant

Code

#include <bits/stdc++.h>
#define ri register int
#define N 1000003
using namespace std;

typedef long long LL;
int n,m,sx,sy;
int a[1003][1003];
int heng[1003],zong[1003];
int dis[N];
LL cnt,ans,tmp1,tmp2,tmpnow;


int main(){
    freopen("travel.in","r",stdin);
    freopen("travel.out","w",stdout);
    scanf("%d%d",&n,&m);
    char ch;
    for(ri i=1;i<=n;++i){
        for(ri j=1;j<=m;++j){
            while(ch=getchar(),(ch!='.')&&(ch!='X'));
            if(ch=='.'){
                a[i][j]=1;
                ++cnt;
            }
            else {
                heng[i]=j;//每一行每一列都只会有一个X 
                zong[j]=i;
            }
        }
    }
    for(ri i=1;i<n;++i){//对x坐标上的曼哈顿距离之和
        tmp1=(heng[i]) ? m-1 : m;
        for(ri j=i+1;j<=n;++j){
            tmp2=(heng[j]) ? m-1 : m;
            ans+=2*tmp1*tmp2*abs(j-i);
        }
    }
    for(ri i=1;i<m;++i){//对y坐标上的曼哈顿距离之和 
        tmp1=(zong[i]!=0) ? n-1 : n;
        for(ri j=i+1;j<=m;++j){
            tmp2=(zong[j]!=0) ? n-1 : n;
            ans+=2*tmp1*tmp2*abs(j-i);
        }
    }
    
    for(ri i=1;i<=m;++i){//跑列 ，找到那些需要+2（思考为什么？）的点对（哪些？） 
        if(!zong[i])continue;
        tmpnow=n-zong[i];
        ans+=2*2*(n-zong[i])*(zong[i]-1);
        int j=i+1;
        while(zong[j]&&zong[j]<zong[j-1]&&j<=m){
            tmp2=zong[j]-1;
            ans+=2*2*tmpnow*tmp2; 
            ++j; 
        } 
        
        j=i-1;
        while(zong[j]&&zong[j]<zong[j+1]&&j>=1){
            tmp1=zong[j]-1;
            ans+=2*2*tmpnow*tmp1; 
            --j;
        } 
    }
    
    for(ri i=1;i<=n;++i){//跑行 
        if(!heng[i])continue;
        ans+=2*2*(m-heng[i])*(heng[i]-1);
        tmpnow=m-heng[i];
        int j=i+1;
        while(heng[j]&&heng[j]<heng[j-1]&&j<=n){
            tmp2=heng[j]-1;
            ans+=2*2*tmpnow*tmp2; 
            ++j; 
        } 
        
        j=i-1;
        while(heng[j]&&heng[j]<heng[j+1]&&j>=1){
            tmp1=heng[j]-1;
            ans+=2*2*tmpnow*tmp1; 
            --j;
        } 
    }
    printf("%.4lf\n",(double)ans/cnt/cnt);
    return 0;
}

LG2824 HEOI2016 / TJOI2016 sort

Correct

1. Half of this location where the number of possible range
   
2. check: mid ratio greater than or equal to 1, otherwise set to 0, in ascending order corresponding to the number of statistics sections 1, 1 is assigned the back section, the front values ​​0

Thinking

1. Wonderful segment tree
   
2. See the changes format, that this is a data structure that, should think of tree line
   
3. But not the sort Maintenance
   
4. Expect to maintain approximately 01 strings, inaccurate sort, it is not accurate the Check, the Check essence without seeking answers to normal Jieci how much, only the possibility of demand