CodeForces 961E Tufurama (静态主席树做法)

题目链接:http://codeforces.com/problemset/problem/961/E

题目大意

给定一个序列a,
问有多少个二元组(x,y)满足
a[x]>=y&&a[y]>=x,x小于y,
统计这样的二元组个数。

题目分析 

明显构建主席树其权值总空间大小是n,
那么我们按序扫描,对于当前值x,
我们需要统计1到x区间中不小于位置i的个数,
注意x要小于i,所以我们取min(x,i-1)保证计数不会重复,
对于其查询不小于x 的个数,对其rt[min(x,i-1)]线段树进行区间统计
即可,因为这颗线段树保存的是前缀和性质。
本人在做题时遇到个想不通的地方就是:一加build初始化就错在
第七个样例,不加就对了(虽然不加确实没影响。。。)但
按照套路应该是加上无妨的。。有大佬看出来了麻烦留言交流下哈T_T...

#include<bits/stdc++.h>
using namespace std;

#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define ll long long

#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define root l,r,rt
#define mst(a,b) memset((a),(b),sizeof(a))
#define pii pair<int,int>
#define fi first
#define se second
#define mk(x,y) make_pair(x,y)
const int mod=1e9+7;
const int maxn=2e5+10;
const int ub=1e6;
const int N=1e9;
ll powmod(ll x,ll y){ll t; for(t=1;y;y>>=1,x=x*x%mod) if(y&1) t=t*x%mod; return t;}
ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
/*
题目大意:
给定一个序列a,
问有多少个二元组(x,y)满足
a[x]>=y&&a[y]>=x,x小于y,
统计这样的二元组个数。

题目分析:
明显构建主席树其权值总空间大小是n,
那么我们按序扫描,对于当前值x,
我们需要统计1到x区间中不小于位置i的个数,
注意x要小于i,所以我们取min(x,i-1)保证计数不会重复,
对于其查询不小于x 的个数,对其rt[min(x,i-1)]线段树进行区间统计
即可,因为这颗线段树保存的是前缀和性质。
本人在做题时遇到个想不通的地方就是:一加build初始化就错在
第七个样例,不加就对了(虽然不加确实没影响。。。)但
按照套路应该是加上无妨的。。有大佬看出来了麻烦留言交流下哈T_T...
*/
int a[maxn],n;
ll ans=0;
int rt[maxn],ls[maxn*20],rs[maxn*20];
int sum[maxn*20];
int tot=0;
void build(int& o,int l,int r){
    o=++tot;
    sum[o]=0;
    if(l==r) return ;
    int mid=l+r>>1;
    build(ls[o],l,mid);
    build(rs[o],mid+1,r);
}
void update(int& o,int l,int r,int last,int p){
    o=++tot;
    sum[o]=sum[last]+1;
    ls[o]=ls[last],rs[o]=rs[last];
    if(l==r) return ;
    int mid=l+r>>1;
    if(p<=mid) update(ls[o],l,mid,ls[last],p);
    else update(rs[o],mid+1,r,rs[last],p);
}
ll Query(int o,int l,int r,int pl,int pr){
    if(pl<=l&&r<=pr) return sum[o];
    int mid=l+r>>1;
    ll ret=0;
    if(pl<=mid) ret+=Query(ls[o],l,mid,pl,pr);
    if(mid<pr) ret+=Query(rs[o],mid+1,r,pl,pr);
    return ret;
}
int main(){
    scanf("%d",&n);
    ///build(rt[0],1,n);不能build。。。。不知道为什么T_T
    rep(i,1,n+1){
        scanf("%d",&a[i]);
        a[i]=min(a[i],n);
        update(rt[i],1,n,rt[i-1],a[i]);
        int tp=min(a[i],i-1);
        ans+=Query(rt[tp],1,n,i,n);
    }
    printf("%lld\n",ans);
    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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