NOI2003文本编辑器

problem

传送门

Solution

块状链表板子题……

码了一下午，调了一晚上，代码重构了3遍，在终于过了。

还是太菜了。

移动光标的操作直接模拟即可。

插入操作，先将光标所在块分裂成两块，然后直接插入。

删除操作直接将边角料变成新块，然后相互连接。

细节有点多……

第一次打，代码奇丑，而且没有优化空间……

算了，以后在填坑吧……

Code

#include <bits/stdc++.h>

using namespace std;

#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define fst first
#define snd second

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

inline int read(){
    int res = 0, fl = 1;
    char r = getchar();
    for (; !isdigit(r); r = getchar()) if(r == '-') fl = -1;
    for (; isdigit(r); r = getchar()) res = (res << 3) + (res << 1) + r - 48;
    return res * fl;
}
typedef long long LL;
typedef pair<int, int> pii;

const int Maxl = 1024 * 1024 * 2, siz = Maxl / 500, blk = Maxl / siz;
char ch[Maxl + 10];
int cnt;

struct node {
    int lst, nxt, len;
    short c[blk + 10];
    void putout(){ for (int i = 1; i <= len; ++i) putchar(c[i]);}
    void back(int C){ c[++len] = C;}
}B[siz << 6];
void check(){
    for (int id = 1; B[id].nxt; id = B[id].nxt){
        if(B[B[id].nxt].len + B[id].len <= blk){
            for (int i = 1; i <= B[B[id].nxt].len; ++i)
                B[id].back(B[B[id].nxt].c[i]);
            B[id].nxt = B[B[id].nxt].nxt;
            B[B[id].nxt].lst = id;
        }
    }
}
int find(int cur, int &Id){
    for (Id = 1; Id && cur > B[Id].len; Id = B[Id].nxt) cur -= B[Id].len;
    return cur;
}
void MakeBlock(int len,int Lst){
    int num = 1;
    B[++cnt].lst = Lst;
    B[Lst].nxt = cnt;
    for (int i = 1; i <= len; ++i){
        if(num * blk + 1 == i){
            B[cnt].nxt = cnt + 1;
            cnt++, num++;
            B[cnt].lst = cnt - 1;
        }
        B[cnt].back(ch[i]);
    }
}
char tmp[5000];
void Insert(int cur, int len){
    int Id;
    cur = find(cur, Id);
    int Nxt = B[Id].nxt, Lst = B[Id].lst;
    MakeBlock(len, Id);
    for (int i = 1; i <= B[Id].len - cur; ++i) ch[i] = B[Id].c[cur + i];
    MakeBlock(B[Id].len - cur, cnt);
    B[cnt].nxt = Nxt;
    B[Nxt].lst = cnt;
    B[Id].len = cur;
    check();
}
void put_out(int cur,int len){
    int Bid, Eid, ecur;
    ecur = find(cur + len, Eid);
    cur = find(cur, Bid);
    if(Bid == Eid){
        for (int i = cur + 1; i <= ecur; ++i) putchar(B[Bid].c[i]);
        putchar('\n');
        return;
    }
    for (int i = cur + 1; i <= B[Bid].len; ++i)putchar(B[Bid].c[i]);
    for (Bid = B[Bid].nxt; Bid != Eid; Bid = B[Bid].nxt)B[Bid].putout();
    for (int i = 1; i <= ecur; ++i) putchar(B[Bid].c[i]);
    putchar('\n');
}
void Erase(int cur, int len){
    int Bid, Eid, ecur;
    ecur = find(cur + len, Eid);
    cur = find(cur, Bid);
    if(Bid == Eid){
        int xz = 0;
        for (int i = 1; i <= cur; ++i) ch[++xz] = B[Bid].c[i];
        for (int i = ecur + 1; i <= B[Bid].len; ++i) ch[++xz] = B[Bid].c[i];
        for (int i = 1; i <= xz; ++i) B[Bid].c[i] = ch[i];
        B[Bid].len = xz;
        check();
        return ;
    }
    B[Bid].len = cur;
    int xz = 0;
    for (int i = ecur + 1; i <= B[Eid].len; ++i) B[Eid].c[++xz] = B[Eid].c[i];
    B[Eid].len = xz;
    B[Bid].nxt = Eid;
    B[Eid].lst = Bid;
    check();
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in", "r", stdin);
    freopen("a.out", "w", stdout);
#endif
    int t = read(), cur = 0;
    char opt[10];
    MakeBlock(blk, 0);
    while(t--){
        scanf("%s",opt + 1);
        if(opt[1] == 'M') cur = read();
        if(opt[1] == 'P') cur--;
        if(opt[1] == 'N') cur++;
        if(opt[1] == 'D') Erase(cur, read());
        if(opt[1] == 'G') put_out(cur, read());
        if(opt[1] == 'I'){
            int len = read();
            for (int i = 1; i <= len; ++i){
                ch[i] = getchar();
                if(ch[i] < 32 || ch[i] > 126) i--;
            }
            Insert(cur, len);
        }
    }
    return 0;
}