Codeforces 260E

本文介绍了一种通过水平和垂直分割平面,并利用线段树进行计数的算法实现。该算法首先枚举子区域的位置,然后通过绘制水平和竖直线将平面分割成多个固定大小的区域,并对这些区域进行计数。

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首先9!枚举儿子的位置,
1.画出水平的两根线,把平面分成三块,每块的个数是个固定值,
2.画出竖直的两根线,把平面分成三块,每块的个数是个固定值,
3.四根线合起来,计数
也不知道这个代码是复制谁的。。。

#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<queue>
#define inf 1000000005
#define M 40
#define N 100005
#define maxn 300005
#define eps 1e-12
#define zero(a) fabs(a)<eps
#define Min(a,b) ((a)<(b)?(a):(b))
#define Max(a,b) ((a)>(b)?(a):(b))
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define MOD 1000000007
#define lson step<<1
#define rson step<<1|1
#define sqr(a) ((a)*(a))
#define Key_value ch[ch[root][1]][0]
#define test puts("OK");
#define pi acos(-1.0)
#define lowbit(x) ((-(x))&(x))
#define HASH1 1331
#define HASH2 10001
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
struct Set_tree{
    int left,right;
    vector<int>v;
}L[N*4];
struct Point{
    int x,y;
    bool operator<(const Point n)const{
        return x!=n.x?x<n.x:y<n.y;
    }
}p[N];
int n,x[N],y[N];
int a[9],b[9];
double ret_x1,ret_x2,ret_y1,ret_y2;
void Bulid(int step,int l,int r){
    L[step].left=l;
    L[step].right=r;
    L[step].v.clear();
    for(int i=l;i<=r;i++)
        L[step].v.pb(p[i].y);
    sort(L[step].v.begin(),L[step].v.end());
    if(l==r)
        return;
    int m=(l+r)>>1;
    Bulid(lson,l,m);
    Bulid(rson,m+1,r);
}
int Query(int step,int l,int r,int val){
    if(L[step].left==l&&r==L[step].right){
        if(L[step].v.size()==0) return 0;
        if(L[step].v[0]>val) return 0;
        if(L[step].v.back()<=val) return L[step].v.size();
        return (upper_bound(L[step].v.begin(),L[step].v.end(),val)-L[step].v.begin());
    }
    int m=(L[step].left+L[step].right)>>1;
    if(r<=m) return Query(lson,l,r,val);
    else if(l>m) return Query(rson,l,r,val);
    else return Query(lson,l,m,val)+Query(rson,m+1,r,val);
}
bool ok(){
    int x1=b[a[0]]+b[a[1]]+b[a[2]]-1;
    int x2=x1+b[a[3]]+b[a[4]]+b[a[5]];
    int y1=b[a[0]]+b[a[3]]+b[a[6]]-1;
    int y2=y1+b[a[1]]+b[a[4]]+b[a[7]];
    if(x1+1>=n||x[x1]==x[x1+1]) return false;
    if(x2+1>=n||x[x2]==x[x2+1]) return false;
    if(y1+1>=n||y[y1]==y[y1+1]) return false;
    if(y2+1>=n||y[y2]==y[y2+1]) return false;
    if(Query(1,0,x1,y[y1])!=b[a[0]]) return false;
    if(Query(1,0,x1,y[y2])!=b[a[0]]+b[a[1]]) return false;
    if(Query(1,x1+1,x2,y[y1])!=b[a[3]]) return false;
    if(Query(1,x1+1,x2,y[y2])!=b[a[3]]+b[a[4]]) return false;
    ret_x1=(x[x1]+x[x1+1])/2.0;
    ret_x2=(x[x2]+x[x2+1])/2.0;
    ret_y1=(y[y1]+y[y1+1])/2.0;
    ret_y2=(y[y2]+y[y2+1])/2.0;
    return true;
}
int main(){
    //freopen("input.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d%d",&p[i].x,&p[i].y);
            x[i]=p[i].x;y[i]=p[i].y;
        }
        sort(p,p+n);
        sort(x,x+n);
        sort(y,y+n);
        Bulid(1,0,n-1);
        for(int i=0;i<9;i++) scanf("%d",&b[i]);
        for(int i=0;i<9;i++) a[i]=i;
        int t=362880;
        while(t--){
            if(ok()){
                printf("%.1f %.1f\n%.1f %.1f\n",ret_x1,ret_x2,ret_y1,ret_y2);
                break;
            }
            next_permutation(a,a+9);
        }
        if(t<=0) puts("-1");
    }
    return 0;
}
### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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