BZOJ 4864: [BeiJing 2017 Wc]神秘物质 (块状链表/平衡树 )

这就是一道数据结构裸题啊，最大极差就是区间最大值减最小值，最小极差就是相邻两个数差的最小值。然后平衡树splay/treap或者块状链表维护就行了。

第一次自己写块状链表，蛮好写，就是长。。然后就BZOJ rank1了（2019.5.11求不打脸 ）

CODE

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
    char ch; for(;!isdigit(ch=getc()););
    for(res=ch-'0';isdigit(ch=getc());res=res*10+ch-'0');
}
const int Bsz = 500;
const int Cnt = 500;
const int inf = 0x3f3f3f3f;

stack<int>bin;
int n, q, num[MAXN], m;
struct Block{
    int a[Bsz+5], use, nxt;
    int mx, mn, d;
    Block() {
        memset(a, 0, sizeof a);
        use = nxt = 0;
        mx = 0, mn = d = inf;
    }
    inline void Clear() {
        memset(a, 0, sizeof a);
        use = nxt = 0;
        mx = 0, mn = d = inf;
    }
    inline void getmax() {
        mx = 0, mn = d = inf;
        for(int i = 1; i <= use; ++i) {
            mx = max(mx, a[i]),
            mn = min(mn, a[i]);
            if(i > 1) d = min(d, abs(a[i]-a[i-1]));
        }
    }
}b[Cnt+5];
inline int Query1(int x, int y) {
    int l = 1;
    while(x > b[l].use) x -= b[l].use, y -= b[l].use, l = b[l].nxt;
    int r = l;
    while(y > b[r].use) y -= b[r].use, r = b[r].nxt;
    int Mx = 0, Mn = inf;
    if(l == r) {
        for(int i = x; i <= y; ++i)
            Mx = max(Mx, b[l].a[i]),
            Mn = min(Mn, b[l].a[i]);
        return Mx-Mn;
    }
    for(int i = x; i <= b[l].use; ++i)
        Mx = max(Mx, b[l].a[i]),
        Mn = min(Mn, b[l].a[i]);
    for(int i = 1; i <= y; ++i)
        Mx = max(Mx, b[r].a[i]),
        Mn = min(Mn, b[r].a[i]);
    for(int i = b[l].nxt; i != r; i = b[i].nxt)
        Mx = max(Mx, b[i].mx),
        Mn = min(Mn, b[i].mn);
    return Mx-Mn;
}
inline int Query2(int x, int y) {
    int l = 1;
    while(x > b[l].use) x -= b[l].use, y -= b[l].use, l = b[l].nxt;
    int r = l;
    while(y > b[r].use) y -= b[r].use, r = b[r].nxt;
    int re = inf;
    if(l == r) {
        for(int i = x; i < y; ++i)
            re = min(re, abs(b[l].a[i]-b[l].a[i+1]));
        return re;
    }
    for(int i = x; i < b[l].use; ++i)
        re = min(re, abs(b[l].a[i]-b[l].a[i+1]));
    for(int i = 1; i < y; ++i)
        re = min(re, abs(b[r].a[i]-b[r].a[i+1]));
    for(int i = b[l].nxt; i != r; i = b[i].nxt)
        re = min(re, b[i].d);
    for(int i = l; i != r; i = b[i].nxt)
        re = min(re, abs(b[i].a[b[i].use]-b[b[i].nxt].a[1]));
    return re;
}
inline int Newnode() {
    int re;
    if(!bin.empty()) re = bin.top(), bin.pop();
    else re = ++m;
    b[re].Clear();
    return re;
}
inline void Merge(int x, int val) {
    int l = 1, pre = 0;
    while(x > b[l].use) x -= b[l].use, pre = l, l = b[l].nxt;
    if(x == b[l].use) {
        int tmp = b[l].nxt;
        b[tmp].a[1] = val;
        b[tmp].getmax();

        if(--b[l].use) b[l].getmax();
        else b[pre].nxt = b[l].nxt, bin.push(l);
    }
    else {
        b[l].a[x+1] = val;
        --b[l].use;
        for(int i = x; i <= b[l].use; ++i)
            b[l].a[i] = b[l].a[i+1];
        b[l].getmax();
    }
}
inline void Insert(int x, int val) {
    int l = 1, pre = 0;
    while(x > b[l].use) x -= b[l].use, pre = l, l = b[l].nxt;
    ++b[l].use;
    for(int i = b[l].use; i > x+1; --i) b[l].a[i] = b[l].a[i-1];
    b[l].a[x+1] = val;
    if(b[l].use > Bsz) {
        int tmp = Newnode();
        b[tmp].nxt = b[l].nxt; b[l].nxt = tmp;
        b[tmp].use = b[l].use/2;
        b[l].use -= b[tmp].use;
        for(int i = 1; i <= b[tmp].use; ++i)
            b[tmp].a[i] = b[l].a[b[l].use+i];
        b[l].getmax(), b[tmp].getmax();
    }
    else b[l].getmax();
}
int main () {
	read(n), read(q);
	for(int i = 1; i <= n; ++i)
        read(num[i]);
    for(int i = 1; i*Bsz <= n; ++i) {
        b[++m].use = Bsz; b[m-1].nxt = m;
        for(int j = 1; j <= b[m].use; ++j)
            b[m].a[j] = num[(m-1)*Bsz + j];
        b[m].getmax();
    }
    if(n%Bsz) {
        b[++m].use = n%Bsz; b[m-1].nxt = m;
        for(int j = 1; j <= b[m].use; ++j)
            b[m].a[j] = num[(m-1)*Bsz + j];
        b[m].getmax();
    }
    char ch;
    int x, y;
	while(q--) {
        while(!isalpha(ch=getc()));
        ch=getc();
        read(x), read(y);
        switch(ch) {
            case 'a':
                printf("%d
", Query1(x, y));
            break;

            case 'i':
                printf("%d
", Query2(x, y));
            break;

            case 'n':
                Insert(x, y);
            break;

            case 'e':
                Merge(x, y);
            break;
        }
	}
}