- 1#include 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
- }
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