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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 = 0;
    while(c < '0' || c > '9') {
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + c - 48;
        c = getchar();
    }
    return;
}

const int N = 1010, M = 100010;

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[10];

struct ROW {
    int a[N], fa[N];
    ROW() {
        for(int i = 1; i < N; i++) {
            fa[i] = i;
            a[i] = 1;
        }
    }
    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 - 1);
            p = fa[p];
        }
        return;
    }
}row1[N], row2[N];

int main() {
    
    read(n);
    read(k);
    read(m);
    
    for(int i = 1; i <= m; i++) {
        scanf("%s", s);
        if(s[0] == 'P') { // print
            fa[i] = i - 1;
            a[i].opt = 1;
            read(a[i].c);
            read(a[i].x);
            read(a[i].y);
            read(a[i].xx);
            read(a[i].yy);
        }
        else if(s[0] == 'S') { // save
            fa[i] = i - 1;
            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 = 1; e <= top; e++) {
        int i = path[e];
        if(!a[i].opt) {
            continue;
        }
        
        for(int j = a[i].x + 1; j <= a[i].xx + 1; j++) {
            //printf("    j = %d  c = %d 
", j, a[i].c);
            if((a[i].x - 1 + a[i].y) & 1) {
                row1[j].change(a[i].c, a[i].y + 1, a[i].yy + 1);
            }
            else {
                row2[j].change(a[i].c, a[i].y + 1, a[i].yy + 1);
            }
        }
    }

    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            if((i + j) & 1) {
                printf("%d ", row2[i].a[j]);
            }
            else {
                printf("%d ", row1[i].a[j]);
            }
        }
        puts("");
    }
    return 0;
}

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 = 0;
    while(c < '0' || c > '9') {
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + c - 48;
        c = getchar();
    }
    return;
}

const int N = 1010, M = 100010;

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[10];
int sa[N * N * 4], sb[N * N * 4], sc[N * N * 4], sd[N * N * 4];
int tag1[N * N * 4], tag2[N * N * 4], G[N][N];

void build(int x, int y, int xx, int yy, int o) {
    if(x == xx && y == yy) {
        tag1[o] = 1;
        return;
    }
    int mx = (x + xx) >> 1;
    int my = (y + yy) >> 1;
    
    sa[o] = ++num;
    build(x, y, mx, my, num);
    
    if(yy > y) {
        sb[o] = ++num;
        build(x, my + 1, mx, yy, num);
    }
    if(xx > x) {
        sc[o] = ++num;
        build(mx + 1, y, xx, my, num);
    }
    
    if(yy > y && xx > x) {
        sd[o] = ++num;
        build(mx +1, my + 1, xx, yy, num);
    }
    return;
}

inline void pushdown(int x, int y, int xx, int yy, int o) {
    int mx = (x + xx) >> 1;
    int my = (y + yy) >> 1;
    if(tag1[o]) {
        int t = tag1[o];
        if(sa[o]) {
            tag1[sa[o]] = t;
        }
        if(sb[o]) {
            if((my - y + 1) & 1) {
                tag2[sb[o]] = t;
            }
            else {
                tag1[sb[o]] = t;
            }
        }
        if(sc[o]) {
            if((mx - x + 1) & 1) {
                tag2[sc[o]] = t;
            }
            else {
                tag1[sc[o]] = t;
            }
        }
        if(sd[o]) {
            if((mx - x + my - y + 1) & 1) {
                tag1[sd[o]] = t;
            }
            else {
                tag2[sd[o]] = t;
            }
        }
        tag1[o] = 0;
    }
    if(tag2[o]) {
        int t = tag2[o];
        if(sa[o]) {
            tag2[sa[o]] = t;
        }
        if(sb[o]) {
            if((my - y + 1) & 1) {
                tag1[sb[o]] = t;
            }
            else {
                tag2[sb[o]] = t;
            }
        }
        if(sc[o]) {
            if((mx - x + 1) & 1) {
                tag1[sc[o]] = t;
            }
            else {
                tag2[sc[o]] = t;
            }
        }
        if(sd[o]) {
            if((mx - x + my - y + 1) & 1) {
                tag2[sd[o]] = t;
            }
            else {
                tag1[sd[o]] = t;
            }
        }
        tag2[o] = 0;
    }
    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) >> 1;
    int my = (y + yy) >> 1;
    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 + 1) & 1) {
            change(v, f ^ 1, x, my + 1, mx, yy, sb[o]);
        }
        else {
            change(v, f, x, my + 1, mx, yy, sb[o]);
        }
    }
    if(mx < XX && Y <= my) { // sc
        if((mx - x + 1) & 1) {
            change(v, f ^ 1, mx + 1, y, xx, my, sc[o]);
        }
        else {
            change(v, f, mx + 1, y, xx, my, sc[o]);
        }
    }
    if(mx < XX && my < YY) { // sd
        if((mx - x + my - y + 1) & 1) {
            change(v, f, mx + 1, my + 1, xx, yy, sd[o]);
        }
        else {
            change(v, f ^ 1, mx + 1, my + 1, 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) >> 1;
    int my = (y + yy) >> 1;
    pushdown(x, y, xx, yy, o);
    ed(x, y, mx, my, sa[o]);
    if(yy > y) {
        ed(x, my + 1, mx, yy, sb[o]);
    }
    if(xx > x) {
        ed(mx + 1, y, xx, my, sc[o]);
    }
    if(xx > x && yy > y) {
        ed(mx + 1, my + 1, xx, yy, sd[o]);
    }
    return;
}

int main() {
    
    read(n);
    read(k);
    read(m);
    
    for(int i = 1; i <= m; i++) {
        scanf("%s", s);
        if(s[0] == 'P') { // print
            fa[i] = i - 1;
            a[i].opt = 1;
            read(a[i].c);
            read(a[i].x);
            read(a[i].y);
            read(a[i].xx);
            read(a[i].yy);
        }
        else if(s[0] == 'S') { // save
            fa[i] = i - 1;
            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 + 1, path + top + 1);
    num = 1;
    build(1, 1, n, n, 1);
    for(int e = 1; 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 - 1 + a[i].y) & 1) {
            change(a[i].c, 1, 1, 1, n, n, 1);
        }
        else {
            change(a[i].c, 0, 1, 1, n, n, 1);
        }
    }
    
    
    ed(1, 1, n, n, 1);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            printf("%d ", G[i][j]);
        }
        puts("");
    }
    return 0;
}

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？