ural 1126【單調隊列基礎】

http://acm.timus.ru/problem.aspx?space=1&num=1126

雖說是簡單的單調隊列，但是足足折磨了我大半天！因為沒接觸過單調隊列，所以剛開始時束手無策，可以用線段樹，但你知道線段樹的代碼量。。。

題意是給出一個數列，給出一個n，問你每n個連續的數中最大的是多少。

具體做法參考我部落格這裡。

代碼實現：本人用了STL的deque

#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <string.h>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>using namespace std;int lowbit(int t){return t&(-t);}int countbit(int t){return (t==0)?0:(1+countbit(t&(t-1)));}int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a>b?b:a;}#define LL __int64#define pi acos(-1)#define N  100010#define INF INT_MAX#define eps 1e-8#define FRE freopen("a.txt","r",stdin)struct node{    int num;    int id;};deque<node> de;int main(){   // FRE;    int n;    scanf("%d",&n);    int cnt=1;    node p;    de.clear();    while(scanf("%d",&p.num)&&p.num!=-1)    {        p.id=cnt;        while(!de.empty()&&de.front().id<cnt-n+1 )            {de.pop_front();}        while(!de.empty()&&de.back().num<p.num)            {de.pop_back();}        cnt++;        de.push_back(p);        if(cnt>n)        printf("%d\n",de.front().num);    }    return 0;}