代码源#452序列操作

Posted 2022-03-05 yeah17981

序列操作 - 题目 - Daimayuan Online Judge

标记一下每个数最后一次进行操作1是什么时候，并找出此操作往后最大的2操作

往后最大用后缀最大值求得

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
int a[maxn],a1[maxn][2],a2[maxn];
int main()

		std::ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	int n,q;
	cin>>n>>q;
	for(int i=1;i<=n;i++)
	
		cin>>a[i];
	
	int flag,x,y,cnt=0;
	for(int i=1;i<=q;i++)
	
		cin>>flag;
		if(flag==1)
		
			cin>>x>>y;
			a1[x][0]=y;
			a1[x][1]=i;
		
		else
		
			cin>>a2[i];
		
	
	int maxx=a2[q];
	for(int i=q-1;i>=0;i--)
	
		if(maxx<a2[i])
		
			maxx=a2[i];
		
		else
		
			a2[i]=maxx;
		
	
	for(int i=1;i<=n;i++)
	
		if(a1[i][0]!=0)
		
			a[i]=a1[i][0];
			
		
		if(a2[a1[i][1]]>a[i])
			
				a[i]=a2[a1[i][1]];
			
	
	for(int i=1;i<=n;i++)
	
		cout<<a[i]<<" ";
	
  

又是买一送一

​​​​​​Dis - 题目 - Daimayuan Online Judge

给一个树，求俩点uv之间最短路的异或和

u到根节点的异或和 异或 v到根节点的异或和 异或 最近公共祖先

用倍增（问就是好写）

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<string.h>
#include<string>
#include<cstring>
#include<map>
#include<unordered_map>
#include<queue>
#include<set>
#include<stdio.h>
#include<stack>
using namespace std;
typedef long long LL;
typedef long double LD;
const int maxn = 3e5;
int fa[maxn][33], depth[maxn],a[maxn];
int n, m,u, v;
vector<int > vec[maxn];
int yihuo[maxn]; //根节点到此点的异或和


void dfs(int u, int pre, int d)  //预处理出每个节点的深度及父亲节点

    fa[u][0] = pre;
    depth[u] = d;
    if(pre!=-1)
	yihuo[u] = yihuo[pre]^a[u];
	else
	yihuo[u]=a[u];
    for (int i = 0; i < vec[u].size(); i++)
    
        int v = vec[u][i];
        if (v != pre)
        
            
            dfs(v, u, d + 1);
        
    
    


void init()                    //预处理出每个节点往上走2^k所到的节点，超过根节点记为-1

    dfs(1, -1, 0);              //root为根节点
    for (int j = 0; (1 << (j + 1)) < n; j++)   //n为节点数目
        for (int i = 0; i < n; i++)
        
            if (fa[i][j] < 0) fa[i][j + 1] = -1;
            else fa[i][j + 1] = fa[fa[i][j]][j];
        


int LCA(int u, int v)

    if (depth[u] > depth[v]) swap(u, v);
    int temp = depth[v] - depth[u];
    for (int i = 0; (1 << i) <= temp; i++)      //使u,v在同一深度
    
        if ((1 << i) & temp)
            v = fa[v][i];
    
    if (v == u) return u;
    for (int i = (int)log2(n * 1.0); i >= 0; i--)  //两个节点一起往上走
    
    	if(depth[fa[u][i]]==0||fa[u][i]==fa[v][i])  continue;
        if (fa[u][i] != fa[v][i])
        
            u = fa[u][i];
            v = fa[v][i];
        
    
    return fa[u][0];

int main()

		std::ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
	
		cin >> a[i];
	
	for (int i = 1; i <= n-1; i++)
	
		cin >> v >> u;
		vec[v].push_back(u);
		vec[u].push_back(v);
	
    init();
    for (int i = 1; i <= m; i++)
    
        cin >> u >> v;
        cout << (yihuo[u] ^ yihuo[v] ^ a[LCA(u, v)]) << "\\n";