42 lines
1.6 KiB
Plaintext
42 lines
1.6 KiB
Plaintext
//稀疏表 Sparse Table
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//一维RMQ 预处理O(nlogn) 查询O(1)
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//返回最大值 下标从1开始 修改即可返回下标
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int p[N] = { -1}, dp[N][20];
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void initRMQ() {
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for (int i = 1; i <= n; i++) { p[i] = p[i >> 1] + 1; dp[i][0] = a[i]; }
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for (int j = 1; j <= p[n]; j++) {
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for (int i = 1; i + (1 << j) - 1 <= n; i++) {
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dp[i][j] = dp[i][j - 1] > dp[i + (1 << (j - 1))][j - 1] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
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}
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}
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}
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inline int query(int l, int r) {
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if (l > r) { swap(l, r); }
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int k = p[r - l + 1];
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return dp[l][k] > dp[r - (1 << k) + 1][k] ? dp[l][k] : dp[r - (1 << k) + 1][k];
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}
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//二维RMQ 预处理O(nmlognlogm) 查询O(1) 下标从1开始
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int p[N] = { -1}, dp[N][N][9][9];
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void initRMQ() {
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for (int i = 1; i <= n; i++) {
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p[i] = p[i >> 1] + 1;
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for (int j = 1; j <= n; j++) { dp[i][j][0][0] = a[i][j]; }
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}
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for (int ii = 0; ii <= p[n]; ii++) {
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for (int jj = 0; jj <= p[m]; jj++) {
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if (ii + jj == 0) { continue; }
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for (int i = 1; i + (1 << ii) - 1 <= n; i++) {
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for (int j = 1; j + (1 << jj) - 1 <= m; j++) {
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if (ii) { dp[i][j][ii][jj] = max(dp[i][j][ii - 1][jj], dp[i + (1 << (ii - 1))][j][ii - 1][jj]); }
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else { dp[i][j][ii][jj] = max(dp[i][j][ii][jj - 1], dp[i][j + (1 << (jj - 1))][ii][jj - 1]); }
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}
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}
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}
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}
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}
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inline int query(int x1, int y1, int x2, int y2) {
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if (x1 > x2) { swap(x1, x2); } if (y1 > y2) { swap(y1, y2); }
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int k1 = p[x2 - x1 + 1], k2 = p[y2 - y1 + 1]; x2 -= (1 << k1) - 1; y2 -= (1 << k2) - 1;
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return max(max(dp[x1][y1][k1][k2], dp[x1][y2][k1][k2]), max(dp[x2][y1][k1][k2], dp[x2][y2][k1][k2]));
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}
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