Tunnel Warfare HDU 1540 + maximum and minimum range of merger

Tunnel Warfare HDU 1540 + maximum and minimum range of merger

The meaning of problems

D x is the destruction of this point, Q x the number of queries is the longest consecutive points where the x, R is the last point of failure recovery.

Problem solution ideas

Reference chiefs blog

Here clever use of the minimum to the maximum to find the interval. The line is a detailed solution to a problem big brother, really wonderful ah.

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#define ls (rt<<1)
#define rs (rt<<1|1)
#define mid ((t[rt].l+t[rt].r)>>1)
using namespace std;
const int maxn=5e4+7;
struct node{
    int l, r, maxx, minn; //维护区间内的最大最小值
}t[maxn<<2];
stack<int> des;
int n, m;
void pushup(int rt)
{
    t[rt].maxx=max(t[ls].maxx, t[rs].maxx);
    t[rt].minn=min(t[ls].minn, t[rs].minn);
}
void build(int rt, int l, int r)
{
    t[rt].l=l;
    t[rt].r=r;
    if(l==r)
    {
        t[rt].maxx=0; //最大值默认都是0
        t[rt].minn=n+1; //最小值默认都是n+1
        return ;
    }
    build(ls, l, mid);
    build(rs, mid+1, r);
    //pushup(rt);
    t[rt].maxx=max(t[ls].maxx, t[rs].maxx);
    t[rt].minn=min(t[ls].minn, t[rs].minn);
}
void destory(int rt, int x)
{
    if(t[rt].l==t[rt].r)
    {
        t[rt].maxx=t[rt].l;
        t[rt].minn=t[rt].l;
        return ;
    }
    if(x<=mid) destory(ls, x);
    else destory(rs, x);
    //pushup(rt);
    t[rt].maxx=max(t[ls].maxx, t[rs].maxx);
    t[rt].minn=min(t[ls].minn, t[rs].minn); 
}
void rebuild(int rt, int x)
{
    if(t[rt].l==t[rt].r)
    {
        t[rt].maxx=0;
        t[rt].minn=n+1;
        return ;
    }
    if(x<=mid) rebuild(ls, x);
    else rebuild(rs, x);
    //pushup(rt);
    t[rt].maxx=max(t[ls].maxx, t[rs].maxx);
    t[rt].minn=min(t[ls].minn, t[rs].minn);
}
int query_max(int rt, int l, int r)
{
    if(l<=t[rt].l && t[rt].r <= r)
    {
        return t[rt].maxx;
    }
    int ans=0;
    if(l<=mid) ans=max(ans, query_max(ls, l, r));
    if(r>mid) ans=max(ans, query_max(rs, l, r));
    return ans; 
}
int query_min(int rt, int l, int r)
{
    if(l<=t[rt].l && t[rt].r <= r)
    {
        return t[rt].minn;
    }
    int ans=0x3f3f3f3f;
    if(l<=mid) ans=min(ans, query_min(ls, l, r));
    if(r>mid) ans=min(ans, query_min(rs, l, r));
    return ans;
}
int main()
{
    char op[3];
    int x;
    while( scanf("%d%d", &n, &m)!=EOF)
    {
        while(!des.empty()) des.pop();
        build(1, 1, n);
        for(int i=1; i<=m; i++)
        {
            scanf("%s", op);
            if(op[0]=='D')
            {
                scanf("%d", &x);
                des.push(x);
                destory(1, x);
            }
            else if(op[0]=='Q')
            {
                scanf("%d", &x);
                int maxx=query_max(1, 1, x);
                int minn=query_min(1, x, n);
                if(maxx==minn)
                    printf("0\n");
                else printf("%d\n", minn-maxx-1);
            }
            else {
                x=des.top();
                des.pop();
                rebuild(1, x);
            }
        }
    }
    return 0;
 }