wuyvle的博客

#include <iostream>#include<stdlib.h>using namespace std;typedef struct{ int a[100]; int len; int maxz=100;} node;typedef node* list;void _insert(list L){ int x; cin>>x; while(x!=-1) { L->a[L

#include<bits/stdc++.h>using namespace std;const int maxm=1010;const int maxn=1010;const int inf=0x3f3f3f3f;int head[maxn],vis[maxn][maxn],dis[maxn][maxn];int cnt,n,m;struct node{ int to,next,l;}e[maxm*2];void add(int x,int y,int l){.

这个题着实折磨#include<stdio.h>#include<iostream>#include<string.h>#define INF 1e14 #define LL long longusing namespace std;char str[55];LL dp[55][55],cost_add[30],cost[30],cost_err[30],C[30][30];//变换表 //dp[i][j]代表i~j段对称的最小花费 void i.

首先知道p[i]≥1，所以不存在p[i]大于等于max(a,b)用map<ll,ll>mp记录下每个p[i]出现的下标，每次找到b-p[i]或a-p[i]都进行一次连接#include <bits/stdc++.h>using namespace std;const int maxn = 1e5+5;int par[maxn];int a[maxn];int n;map<int,int> mp;void init(){ for (int i=0.

那么只要原来的蚯蚓具有单调性，每次只需要找这三个队列的最大值即可至于每次都要增加的长度，我们可以先不加上，等到取出来的时候在加 (i - 1) * q(i−1)∗q, 放回去的时候注意要减去 i * qi∗q#include<bits/stdc++.h>typedef long long ll;using namespace std;const int N = 2e5 + 5;const int mod = 1e9 + 7;queue<ll> q[3];ll a.

#include <iostream>using namespace std;#define INFINITY 100000000struct gnode{ int weight; int expenses;}G[500][500];int nv, ne;int sta, enk;int visit[500] = { 0 };int dist[500], cost[500];void build(){//建立一个表 int v1, v2, w, ex; fo.

#include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> #define INF 0x3f3f3f3f using namespace std; int dis[102],map[110][110]; int n,m; void floyd(){ int i,k,j; for.

#include"iostream"#include"string.h"#include"stack"#include"vector"#include"algorithm"using namespace std;typedef struct node{ int to; int cost;}node;int in[101];int dis[101];stack<int>q;vector<node>t[101];int cnt = 0;void topo.

#include <stdio.h>#include <malloc.h>#define ElemType inttypedef struct BiTNode { ElemType Data; struct BiTNode *Left; struct BiTNode *Right;}*AVLTree;AVLTree SingleLeftRotation(AVLTree T) {//左单旋 AVLTree B=T->Left; ...