Please suggest some problems related to priority queue concept as Codeforces data structure tag is quite bigger . Thanks and regards...priorityqueue, problems, request -11 nipul1 6 years ago 1 Comments (1) Write comment? Sturdy 6 years ago, # | -9 I don't understand why people use...
http://codeforces.com/contest/802/problem/B (4) //Priority by "next"↵↵10850 UVA (4) //Queue a brute-force↵↵http://codeforces.com/contest/813/problem/A (1) //Zahrivacka pro prvaky↵↵10716 UVA (4) //NICE — always find closest pair↵↵http://codeforces....
priority_queue <int,vector<int>,greater<int> > q; 代码: 1#include <iostream>2#include <vector>3#include <queue>4#definemax_n 2000055usingnamespacestd;6vector<pair<int,int> >vec[max_n];7structnode8{9intid;10inttime;11friendbooloperator<(node a,node b)12{13returna.time<b.time;14}...
#include<bits/stdc++.h>usingnamespacestd; priority_queue<int, vector<int>, greater<int>> pq;constintN =1e5+10;inthead[N], vis[N], pnt[N <<1], nxt[N <<1], cnt, n, m, x, y; vector<int> ans;voidadd(intx,inty){ pnt[cnt] = y; nxt[cnt] = head[x]; head[x] = cnt...
思路II 同思路I,使得差分数组非负.将差分数组的值加入小根堆中,倒序枚举每次操作i.每次取出最小的差分值,令其+=i后放回小根堆中. 该思路的解的存在性和正确性的证明建立在思路I构造的方案上. 代码II typedefpair<int,int>pii;voidsolve(){intn;cin>>n;vector<int>a(n+1);priority_queue<pii,vector<pi...
priority_queue<hh> q; dis[s]=0; len[s]=0; q.push(hh(dis[s],s)); while(!q.empty()){ hh tmp; tmp=q.top(); q.pop(); int u=tmp.v; for (int i = head[u];~i;i=a[i].nt){ int v=a[i].v; if(dis[v]>dis[u]+a[i].w||(dis[v]==dis[u]+a[i].w&&len[v...
#include<bits/stdc++.h>using namespace std;int n,t,a[3000000];struct node{int first,second;bool operator<(constnode&b)const{if(first==b.first)returnsecondb.first;}};priority_queue<node>ms;intmain(){cin>>n;for(int i=0;i<n;++i){cin>>a[i];}for(int i=0;i<n;++i){int b;...
priority_queue<int,vector<int>,less<int>>p;//大到小 priority_queue<int,vector<int>,greater<int>>q;//小到大 constllmaxn=100000+5; vector<int>a[maxn];//定义一个动态数组 二维数组也可以吧。 boolvis[maxn];//记录未被遍历的 intmain() ...
priority_queue<node>q; LL a[maxn], n, k, x, flag; void init() { scanf("%lld%lld%lld",&n, &k, &x); while(!q.empty()) q.pop(); flag = 0; for(int i = 1; i<=n; i++) { node e; scanf("%lld",&a[i]); e.v = abs(a[i]); e.s = (a[i]<0); e.pos...
priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<>> pq; pq.emplace(0, 0, 0); for (;;) { auto [d, i, j] = pq.top(); pq.pop(); if (i == n - 1 && j == m - 1) { return d; } if (d > dis[i][j]) { continue; } for (auto& ...