题目链接 Print Check
注意到行数加列数最大值只有几千,那么有效的操作数只有几千,那么把这些有效的操作求出来依次模拟就可以了。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define rep(i,a,b) for(int i(a); i <= (b); ++i) 6 const int N = 100000 + 10; 7 const int Q = 5000 + 10; 8 9 struct node{ 10 int op, x, y; 11 } a[N]; 12 13 int col[Q][Q]; 14 int v[N]; //v[i] = 1 则第i个操作需要被执行 15 int c[3][Q]; 16 int n, m, k, op, x, y; 17 18 int main(){ 19 20 scanf("%d%d%d", &n, &m, &k); 21 rep(i, 1, k){ 22 scanf("%d%d%d", &op, &x, &y); 23 c[op][x] = i; 24 a[i].op = op, a[i].x = x, a[i].y = y; 25 } 26 27 rep(i, 1, n) if (c[1][i]) v[c[1][i]] = 1; 28 rep(i, 1, m) if (c[2][i]) v[c[2][i]] = 1; 29 30 rep(i, 1, k) if (v[i]){ 31 if (a[i].op == 1){ 32 rep(j, 1, m) col[a[i].x][j] = a[i].y; 33 } 34 else{ 35 rep(j, 1, n) col[j][a[i].x] = a[i].y; 36 } 37 } 38 39 rep(i, 1, n){ 40 rep(j, 1, m) printf("%d ", col[i][j]); 41 putchar(10); 42 } 43 44 return 0; 45 46 }