NOIP模拟（10.20）T3 裁剪表格

裁剪表格

题目背景：

10.20 NOIP模拟T3

分析：链表······

当看到题目中的高级数据结构就开始胡思乱想的一定不止我一个人，然而这道题被暴力操的一干二净，然而给平衡树的点暴力秒秒钟卡过，直接就AC了······我也是······（论出题人崩溃的内心）

首先对于30%的数据，很显然的直接swap就好，但是你会发现其实60%的数据，其实暴力也可以很方便的卡过去······（然而加一个register连100%的数据都卡过了······）

Source：

/*created by scarlyw*/#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <algorithm>#include <cctype>#include <set>#include <map>#include <vector>#include <queue>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;}/*template<class T>inline void R(T &x) {static bool iosig;static char c;for (iosig = false, c = read(); !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;}*/const int OUT_LEN = 1024 * 1024;char obuf[OUT_LEN], *oh = obuf;inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;}template<class T>inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}}inline void flush() {fwrite(obuf, 1, oh - obuf, stdout);}///*template<class T>inline void R(T &x) {static bool iosig;static char c;for (iosig = false, c = getchar(); !isdigit(c); c = getchar()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;}//*/const int MAXN = 1000 + 10;int n, m, q;int table[MAXN][MAXN];inline void read_in() {R(n), R(m), R(q);for (register int i = 1; i <= n; ++i)for (register int j = 1; j <= m; ++j)R(table[i][j]);}inline void solve() {register int x1, y1, x2, y2, h, w;while (q--) {R(x1), R(y1), R(x2), R(y2), R(h), R(w);for (register int i = 0; i < h; ++i)for (register int j = 0; j < w; ++j)std::swap(table[x1 + i][y1 + j], table[x2 + i][y2 + j]);}for (register int i = 1; i <= n; ++i) {for (register int j = 1; j <= m; ++j)W(table[i][j]), write_char(' ');write_char('\n');}}int main() {//freopen("table.in", "r", stdin);//freopen("table.out", "w", stdout);read_in();solve();flush();return 0;}

但是出题人表示，本来60%的档是给平衡树的，你对于每一行开一颗平衡树，然后暴力提取区间再接到对应位置即可。

 

我们来讲讲真正的正解，正解是链表······4向链表，往四个方向都维护一下就好，但是实际操作上可以直接只维护两个防线即可，对于每一次操作相当于断掉当前两个矩阵周围一圈向外面的指针，然后依次交换就好，因为矩阵的周长之和和找到对应位置所需要的次数之和是O(n)的所以整体复杂度就是O(nq)的，详见代码

Source：

/*created by scarlyw*/#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <algorithm>#include <cctype>#include <set>#include <map>#include <vector>#include <queue>#include <ctime>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;}///*template<class T>inline void R(T &x) {static bool iosig;static char c;for (iosig = false, c = read(); !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;}//*/const int OUT_LEN = 1024 * 1024;char obuf[OUT_LEN], *oh = obuf;inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;}template<class T>inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}}inline void flush() {fwrite(obuf, 1, oh - obuf, stdout);}/*template<class T>inline void R(T &x) {static bool iosig;static char c;for (iosig = false, c = getchar(); !isdigit(c); c = getchar()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;}//*/const int MAXX = 1000000 + 50000;const int MAXN = 1000 + 10;int n ,m ,q, ind;int num[MAXN][MAXN], f[MAXX][2], v[MAXX];inline void read_in() {R(n), R(m), R(q);for (register int i = 1; i <= n; ++i) for (register int j = 1; j <= m; ++j)R(v[num[i][j] = ++ind]);for (register int i = 0; i <= m + 1; ++i) num[0][i] = ++ind;for (register int i = 0; i <= m + 1; ++i) num[n + 1][i] = ++ind;for (register int i = 1; i <= n; ++i) num[i][0] = ++ind;for (register int i = 1; i <= n; ++i) num[i][m + 1] = ++ind;for (register int i = 0; i <= n; ++i)for (register int j = 0; j <= m; ++j)f[num[i][j]][0] = num[i][j + 1], f[num[i][j]][1] = num[i + 1][j];}inline int get_num(int x, int y) {int pos = num[0][0];for (register int i = 1; i <= x; ++i) pos = f[pos][1];for (register int i = 1; i <= y; ++i) pos = f[pos][0];return pos;} inline void solve() {register int x1, y1, x2, y2, h, w;while (q--) {R(x1), R(y1), R(x2), R(y2), R(h), R(w);int pos1 = get_num(x1 - 1, y1 - 1), pos2 = get_num(x2 - 1, y2 - 1);register int i, temp1, temp2;for (i = 1, temp1 = pos1, temp2 = pos2; i <= w; ++i)std::swap(f[temp1 = f[temp1][0]][1], f[temp2 = f[temp2][0]][1]);for (i = 1; i <= h; ++i)std::swap(f[temp1 = f[temp1][1]][0], f[temp2 = f[temp2][1]][0]);for (i = 1, temp1 = pos1, temp2 = pos2; i <= h; ++i)std::swap(f[temp1 = f[temp1][1]][0], f[temp2 = f[temp2][1]][0]);for (i = 1; i <= w; ++i)std::swap(f[temp1 = f[temp1][0]][1], f[temp2 = f[temp2][0]][1]);}for (register int i = 1, pos = num[0][0]; i <= n; ++i, write_char('\n'))for (register int j = 1, temp = pos = f[pos][1]; j <= m; ++j)W(v[temp = f[temp][0]]), write_char(' ');}int main() {read_in();solve();flush();return 0;}