Tunnel Warfare 线段树 区间合并|最大最小值

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!

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
 */

原文地址：https://www.cnblogs.com/EchoZQN/p/10802782.html