chessboard

题意：n*n的矩阵，m次赋值一个子矩阵为c，最后输出答案。

n<=1e3 m<=1e5

解：倒序处理。

拆行处理。

每行内并查集维护未被赋值的地方。

这样每个地方最多被赋值一次，每次修改要访问n行，时间复杂度是O(n(n + m))

 #include <cstdio>

 inline void read(int &x) {
     char c = getchar();
     x = ;
     while(c < '' || c > '') {
         c = getchar();
     }
     while(c >= '' && c <= '') {
         x = (x << ) + (x << ) + c - ;
         c = getchar();
     }
     return;
 }

 const int N = , M = ;

 struct Node {
     int opt, c, x, y, xx, yy;
 }a[M]; 

 int n, k, m, fa[M], sv[M], num, path[M], top;
 char s[];

 struct ROW {
     int a[N], fa[N];
     ROW() {
         for(int i = ; i < N; i++) {
             fa[i] = i;
             a[i] = ;
         }
     }
     int find(int x) {
         if(x == fa[x]) {
             return x;
         }
         return fa[x] = find(fa[x]);
     }
     inline void change(int v, int l, int r) {
         int p = find(r);
         while(p >= l) {
             a[p] = v;
             fa[p] = find(p - );
             p = fa[p];
         }
         return;
     }
 }row1[N], row2[N];

 int main() {

     read(n);
     read(k);
     read(m);

     for(int i = ; i <= m; i++) {
         scanf("%s", s);
         if(s[] == 'P') { // print
             fa[i] = i - ;
             a[i].opt = ;
             read(a[i].c);
             read(a[i].x);
             read(a[i].y);
             read(a[i].xx);
             read(a[i].yy);
         }
         else if(s[] == 'S') { // save
             fa[i] = i - ;
             sv[++num] = i;
         }
         else { // load
             int x;
             //scanf("%d", &x);
             read(x);
             fa[i] = sv[x];
         }
     }

     int x = m;
     while(x) {
         path[++top] = x;
         x = fa[x];
     }

     for(int e = ; e <= top; e++) {
         int i = path[e];
         if(!a[i].opt) {
             continue;
         }

         for(int j = a[i].x + ; j <= a[i].xx + ; j++) {
             //printf("    j = %d  c = %d \n", j, a[i].c);
             if((a[i].x -  + a[i].y) & ) {
                 row1[j].change(a[i].c, a[i].y + , a[i].yy + );
             }
             else {
                 row2[j].change(a[i].c, a[i].y + , a[i].yy + );
             }
         }
     }

     for(int i = ; i <= n; i++) {
         for(int j = ; j <= n; j++) {
             if((i + j) & ) {
                 printf("%d ", row2[i].a[j]);
             }
             else {
                 printf("%d ", row1[i].a[j]);
             }
         }
         puts("");
     }
     return ;
 }

AC代码

说一下原来的题意：

有个矩阵，每次间隔染色一个子矩阵(类似国际象棋那样，左上角要染)，还要支持存档/读档。最后一次输出。

 #include <cstdio>
 #include <algorithm>

 inline void read(int &x) {
     char c = getchar();
     x = ;
     while(c < '' || c > '') {
         c = getchar();
     }
     while(c >= '' && c <= '') {
         x = (x << ) + (x << ) + c - ;
         c = getchar();
     }
     return;
 }

 const int N = , M = ;

 struct Node {
     int opt, c, x, y, xx, yy;
 }a[M]; 

 int n, k, m, fa[M], sv[M], num, path[M], top;
 char s[];
 int sa[N * N * ], sb[N * N * ], sc[N * N * ], sd[N * N * ];
 int tag1[N * N * ], tag2[N * N * ], G[N][N];

 void build(int x, int y, int xx, int yy, int o) {
     if(x == xx && y == yy) {
         tag1[o] = ;
         return;
     }
     int mx = (x + xx) >> ;
     int my = (y + yy) >> ;

     sa[o] = ++num;
     build(x, y, mx, my, num);

     if(yy > y) {
         sb[o] = ++num;
         build(x, my + , mx, yy, num);
     }
     if(xx > x) {
         sc[o] = ++num;
         build(mx + , y, xx, my, num);
     }

     if(yy > y && xx > x) {
         sd[o] = ++num;
         build(mx +, my + , xx, yy, num);
     }
     return;
 }

 inline void pushdown(int x, int y, int xx, int yy, int o) {
     int mx = (x + xx) >> ;
     int my = (y + yy) >> ;
     if(tag1[o]) {
         int t = tag1[o];
         if(sa[o]) {
             tag1[sa[o]] = t;
         }
         if(sb[o]) {
             if((my - y + ) & ) {
                 tag2[sb[o]] = t;
             }
             else {
                 tag1[sb[o]] = t;
             }
         }
         if(sc[o]) {
             if((mx - x + ) & ) {
                 tag2[sc[o]] = t;
             }
             else {
                 tag1[sc[o]] = t;
             }
         }
         if(sd[o]) {
             if((mx - x + my - y + ) & ) {
                 tag1[sd[o]] = t;
             }
             else {
                 tag2[sd[o]] = t;
             }
         }
         tag1[o] = ;
     }
     if(tag2[o]) {
         int t = tag2[o];
         if(sa[o]) {
             tag2[sa[o]] = t;
         }
         if(sb[o]) {
             if((my - y + ) & ) {
                 tag1[sb[o]] = t;
             }
             else {
                 tag2[sb[o]] = t;
             }
         }
         if(sc[o]) {
             if((mx - x + ) & ) {
                 tag1[sc[o]] = t;
             }
             else {
                 tag2[sc[o]] = t;
             }
         }
         if(sd[o]) {
             if((mx - x + my - y + ) & ) {
                 tag2[sd[o]] = t;
             }
             else {
                 tag1[sd[o]] = t;
             }
         }
         tag2[o] = ;
     }
     return;
 }

 int X, XX, Y, YY;
 inline void change(int v, int f, int x, int y, int xx, int yy, int o) {
     int mx = (x + xx) >> ;
     int my = (y + yy) >> ;
     if(X <= x && xx <= XX) {
         if(Y <= y && yy <= YY) {
             if(f) {
                 tag1[o] = v;
             }
             else {
                 tag2[o] = v;
             }
             return;
         }
     }
     pushdown(x, y, xx, yy, o);
     if(X <= mx && Y <= my) { // sa
         change(v, f, x, y, mx, my, sa[o]);
     }
     if(X <= mx && my < YY) { // sb
         if((my - y + ) & ) {
             change(v, f ^ , x, my + , mx, yy, sb[o]);
         }
         else {
             change(v, f, x, my + , mx, yy, sb[o]);
         }
     }
     if(mx < XX && Y <= my) { // sc
         if((mx - x + ) & ) {
             change(v, f ^ , mx + , y, xx, my, sc[o]);
         }
         else {
             change(v, f, mx + , y, xx, my, sc[o]);
         }
     }
     if(mx < XX && my < YY) { // sd
         if((mx - x + my - y + ) & ) {
             change(v, f, mx + , my + , xx, yy, sd[o]);
         }
         else {
             change(v, f ^ , mx + , my + , xx, yy, sd[o]);
         }
     }
     return;
 }

 void ed(int x, int y, int xx, int yy, int o) {
     if(x == xx && y == yy) {
         G[x][y] = tag1[o];
         return;
     }
     int mx = (x + xx) >> ;
     int my = (y + yy) >> ;
     pushdown(x, y, xx, yy, o);
     ed(x, y, mx, my, sa[o]);
     if(yy > y) {
         ed(x, my + , mx, yy, sb[o]);
     }
     if(xx > x) {
         ed(mx + , y, xx, my, sc[o]);
     }
     if(xx > x && yy > y) {
         ed(mx + , my + , xx, yy, sd[o]);
     }
     return;
 }

 int main() {

     read(n);
     read(k);
     read(m);

     for(int i = ; i <= m; i++) {
         scanf("%s", s);
         if(s[] == 'P') { // print
             fa[i] = i - ;
             a[i].opt = ;
             read(a[i].c);
             read(a[i].x);
             read(a[i].y);
             read(a[i].xx);
             read(a[i].yy);
         }
         else if(s[] == 'S') { // save
             fa[i] = i - ;
             sv[++num] = i;
         }
         else { // load
             int x;
             read(x);
             fa[i] = sv[x];
         }
     }

     int x = m;
     while(x) {
         path[++top] = x;
         x = fa[x];
     }
     std::reverse(path + , path + top + );
     num = ;
     build(, , n, n, );
     for(int e = ; e <= top; e++) {
         int i = path[e];
         if(!a[i].opt) {
             continue;
         }
         a[i].x++;
         a[i].xx++;
         a[i].y++;
         a[i].yy++;
         X = a[i].x;
         Y = a[i].y;
         XX = a[i].xx;
         YY = a[i].yy;
         if((a[i].x -  + a[i].y) & ) {
             change(a[i].c, , , , n, n, );
         }
         else {
             change(a[i].c, , , , n, n, );
         }
     }

     ed(, , n, n, );
     for(int i = ; i <= n; i++) {
         for(int j = ; j <= n; j++) {
             printf("%d ", G[i][j]);
         }
         puts("");
     }
     return ;
 }

四分树

这是考场上写的，四分树每次修改貌似是O(n)的，但是卡成6s......只有75分。

话说我第一次写四分树居然是在考场上，而且一个字符都没调就一次写对了......

时间复杂度到底是多少呀......nm的话应该能过呀？莫非是我常数大？