CodeForces 811B Vladik and Complicated Book

离线，树状数组。

数据范围好像有点小，直接暴力可以过的。

我直接上了$n,Q≤100000$的做法：只需要判断区间上比$x$小的数字有几个即可，可以对询问进行离线操作，从左到右一个一个数字插入到树状数组中。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>
using namespace std;int n,m;
struct X
{int id;int op;int pos;int val;int y;
}s[20010];
int sz;
int ans[10010][2];
int c[10100],a[10100];int lowbit(int x)
{return x&(-x);
}void update(int x)
{while(x<=10000){c[x]++;x = x+lowbit(x);}
}int get(int x)
{int res = 0;while(x>0){res = res + c[x];x = x-lowbit(x);}return res;
}bool cmp(X a,X b)
{return a.pos<b.pos;
}bool cmp2(X a,X b)
{if(a.id == b.id) return a.op < b.op;return a.id < b.id;
}int main()
{scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) scanf("%d",&a[i]);for(int i=1;i<=m;i++){int L,R,v; scanf("%d%d%d",&L,&R,&v);s[sz].id = i;s[sz].op = 0;s[sz].pos = L-1;s[sz].val = a[v];s[sz].y = v;sz++;s[sz].id = i;s[sz].op = 1;s[sz].pos = R;s[sz].val = a[v];s[sz].y = v;sz++;}sort(s,s+sz,cmp);int p = 0;for(int i=0;i<=n;i++){if(i>0) update(a[i]);while(1){if(p<sz && s[p].pos==i){ans[s[p].id][s[p].op] = get(s[p].val-1);p++;}else break;}}sort(s,s+sz,cmp2);int id = 0;for(int i=0;i<sz;i=i+2){id++;int num1 = ans[id][1] - ans[id][0];int num2 = s[i].y - s[i].pos - 1;if(num1 == num2) printf("Yes\n");else printf("No\n");}return 0;
}

转载于:https://www.cnblogs.com/zufezzt/p/6930423.html

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