标签：request efi upd one pac color arm state 前缀
InputThe first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.
There are three different events described in different format shown below:
D x: The x-th village was destroyed.
Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.
R: The village destroyed last was rebuilt.
OutputOutput the answer to each of the Army commanders’ request in order on a separate line.
Sample Input
7 9
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4
Sample Output
1 0 2 4
#include <cstdio>//法一：最大值最小值法，这个就是每一个点，如果这个点没有被摧毁，那就找到这个点最左边的和最右边的
#include <cstdlib>//最大-最小+1.这个就是这个最大连续长度。
#include <queue>//建树，很简单，主要就是query和update。
#include <algorithm>//这个地方的怎么去找一个包含一个数的一个区间的最大最小值呢？
#include <vector>//这个就是从上面往下面查询的过程中，就去找，如果是找最大值就去max，最小值就取min
#include <cstring>//这个要注意建树，这个区间的最大值的意思是，小于等于这个数的最大的被炸了的村庄，这个就说明，开始最大值为0，因为没有任何一个村庄被炸
#include <string>//区间的最小值，意思是大于等于这个数，被炸了的村庄的最小值，开始为n+1.因为没有村庄被炸。
#include <iostream>
#include <stack>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1e5 + 100;
struct node
{
int l, r;
int mx, mn;
}tree[maxn*4];
int n;
void build(int id,int l,int r)
{
tree[id].l = l;
tree[id].r = r;
if(l==r)
{
tree[id].mn = n+1;
tree[id].mx = 0;
return;
}
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
tree[id].mx = max(tree[id << 1].mx, tree[id << 1 | 1].mx);
tree[id].mn = min(tree[id << 1].mn, tree[id << 1 | 1].mn);
}
void update_max(int id,int x,int z)
{
if(tree[id].l==tree[id].r)
{
tree[id].mx = z;
return;
}
int mid = (tree[id].l + tree[id].r) >> 1;
if(x<=mid) update_max(id << 1, x, z);
if (x > mid) update_max(id << 1 | 1, x, z);
tree[id].mx = max(tree[id << 1].mx, tree[id << 1 | 1].mx);
}
void update_min(int id,int x,int z)
{
if(tree[id].l==tree[id].r)
{
tree[id].mn = z;
return;
}
int mid = (tree[id].l + tree[id].r) >> 1;
if (x <= mid) update_min(id << 1, x, z);
if (x > mid) update_min(id << 1 | 1, x, z);
tree[id].mn = min(tree[id << 1].mn, tree[id << 1 | 1].mn);
}
int query_max(int id,int x,int y)
{
int ans = 0;
if(x<=tree[id].l&&y>=tree[id].r)
{
return tree[id].mx;
}
int mid = (tree[id].l + tree[id].r) >> 1;
if (x <= mid) ans = max(ans, query_max(id << 1, x, y));
if (y > mid) ans = max(ans, query_max(id << 1 | 1, x, y));
return ans;
}
int query_min(int id,int x,int y)
{
int ans = inf;
if(x<=tree[id].l&&y>=tree[id].r)
{
return tree[id].mn;
}
int mid = (tree[id].l + tree[id].r) >> 1;
if (x <= mid) ans = min(ans, query_min(id << 1, x, y));
if (y > mid) ans = min(ans, query_min(id << 1 | 1, x, y));
return ans;
}
int main()
{
int m, x;
while(cin >> n >> m)
{
stack<int>sta;
build(1, 1, n);
while(m--)
{
char s[10];
scanf("%s", s);
if(s[0]==‘D‘)
{
cin >> x;
update_max(1, x, x);
update_min(1, x, x);
sta.push(x);
}
if(s[0]==‘R‘)
{
int y = sta.top(); sta.pop();
update_max(1, y, 0);
update_min(1, y, n + 1);
}
if(s[0]==‘Q‘)
{
cin >> x;
int L = query_min(1, x, n + 1);
int R = query_max(1, 0, x);
//printf("%d %d\n", L, R);
if (L == R) printf("0\n");
else printf("%d\n", L - R - 1);
}
}
}
return 0;
}
//法二：区间合并，这个应该更好懂一点，就是维护一下一个区间的前缀后缀长度
//这个更新应该比较简单，
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <string>
#include <iostream>
#include <stack>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1e5 + 100;
struct node
{
int l, r, len;
int max_pre, max_last;
}tree[maxn*4];
void push_up(int id)
{
tree[id].max_pre = tree[id << 1].max_pre;
tree[id].max_last = tree[id << 1 | 1].max_last;
if (tree[id << 1].max_pre == tree[id << 1].len)
{
tree[id].max_pre += tree[id << 1 | 1].max_pre;
}
if(tree[id<<1|1].max_last==tree[id<<1|1].len)
{
tree[id].max_last += tree[id << 1].max_last;
}
}
void build(int id,int l,int r)
{
tree[id].l = l;
tree[id].r = r;
tree[id].len = r - l + 1;
if(l==r)
{
tree[id].max_pre = tree[id].max_last = 1;
return;
}
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
push_up(id);
}
void update(int id,int x,int z)
{
if(tree[id].l==tree[id].r)
{
tree[id].max_pre = z;
tree[id].max_last = z;
return;
}
int mid = (tree[id].l + tree[id].r) >> 1;
if (x <= mid) update(id << 1, x, z);
else update(id << 1 | 1, x, z);
push_up(id);
}
int query_pre(int id,int x,int y)
{
int ans = 0, res = 0;
if(x<=tree[id].l&&y>=tree[id].r)
{
//printf("tree[%d].max_pre=%d\n", id, tree[id].max_pre);
return tree[id].max_pre;
}
//printf("id=%d x=%d y=%d\n", id, x, y);
int mid = (tree[id].l + tree[id].r) >> 1;
if (x <= mid) ans=query_pre(id << 1, x, y);
if (y > mid) res=query_pre(id << 1 | 1, x, y);
//printf("id=%d ans=%d res=%d mid=%d\n",id, ans, res,mid);
if (ans >= mid - x + 1)
{
//printf("tree[%d].max_pre=%d mid=%d x=%d\n",id, tree[id].max_pre, mid, x);
ans += res;
}
return ans;
}
int query_last(int id,int x,int y)
{
int ans = 0, res = 0;
if (x <= tree[id].l&&y >= tree[id].r)
{
//printf("tree[%d].last=%d\n", id, tree[id].max_last);
return tree[id].max_last;
}
//printf("id=%d x=%d y=%d\n", id, x, y);
int mid = (tree[id].l + tree[id].r) >> 1;
if (x <= mid) ans = query_last(id << 1, x, y);
if (y > mid) res = query_last(id << 1 | 1, x, y);
//printf("id=%d mid=%d ans=%d res=%d\n", id, mid,ans, res);
if (res >= y-mid)
{
//printf("tree[%d].max_last=%d mid=%d x=%d\n",id,tree[id].max_last, mid, x);
res += ans;
}
return res;
}
int main()
{
int n, m;
while(scanf("%d %d",&n,&m)!=EOF)
{
stack<int>sta;
build(1, 1, n);
while(m--)
{
char s[10];
scanf("%s", s);
if(s[0]==‘D‘)
{
int x;
cin >> x;
update(1, x, 0);
sta.push(x);
}
if(s[0]==‘R‘)
{
int y = sta.top(); sta.pop();
update(1, y, 1);
}
if(s[0]==‘Q‘)
{
int x;
cin >> x;
int ans = query_pre(1, x, n);
ans += query_last(1, 1, x);
if(ans) printf("%d\n", ans-1);
else printf("0\n");
}
}
}
return 0;
}
/*
7 10
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4
Q 3
*/
标签：request efi upd one pac color arm state 前缀
原文地址：https://www.cnblogs.com/EchoZQN/p/10802782.html
