Submission #159779

#TimeUsernameProblemLanguageResultExecution timeMemory
159779rama_pangWombats (IOI13_wombats)C++14
76 / 100
8135 ms262148 KiB
#include "wombats.h" #include <bits/stdc++.h> using namespace std; using lint = long long; const lint INF = 1e9; /* Use segment tree, each node all C * C answers of interval l...r. For query, it can be answered in O(1) by querying the root, V1 and V2 To merge, done naively is O(C * C * C). However, observe that we can apply Knuth's optimization to make it O(C * C): observe that the grids are planar, thus if will always follow the requirements for Knuth's optimization. Also, O(R * C * C) memory is too large, thus we can decompose the grid into blocks of a certain size, then build a segment tree of those blocks. To update a block, just brute force it. */ lint R, C; lint H[5555][255], V[5555][255]; const lint BLOCK_SIZE = 20; const lint BLOCK_COUNT = (5005 / BLOCK_SIZE); struct segment_tree { struct node { vector<vector<lint>> paths; node(bool upd = false, int block = 0) { paths.assign(C, vector<lint>(C, INF)); if (!upd) return; for (int i = 0; i < C; i++) { paths[i][i] = 0; for (int k = block * BLOCK_SIZE; k < (block + 1) * BLOCK_SIZE; k++) { if (k >= R) continue; for (int j = 1; j < C; j++) { paths[i][j] = min(paths[i][j], paths[i][j - 1] + H[k][j - 1]); } for (int j = C - 2; j >= 0; j--) { paths[i][j] = min(paths[i][j], paths[i][j + 1] + H[k][j]); } for (int j = 0; j < C; j++) { paths[i][j] += V[k][j]; } } } } node merge(node a, node b) { node res; vector<vector<lint>> mid(C, vector<lint>(C, 0)); // for (int len = 0; len < C; len++) { // for (int i = 0; i + len < C; i++) { // int j = i + len; // res.paths[i][j] = INF; // if (len == 0) { // for (int k = 0; k < C; k++) { // if (res.paths[i][j] > a.paths[i][k] + b.paths[k][j]) { // mid[i][j] = k; // res.paths[i][j] = a.paths[i][k] + b.paths[k][j]; // } // } // continue; // } // /* DP Knuth's optimization, computing paths for new interval */ // int le = mid[i][j - 1], ri = mid[i + 1][j]; // for (int k = le; k <= ri; k++) { // if (res.paths[i][j] > a.paths[i][k] + b.paths[k][j]) { // mid[i][j] = k; // res.paths[i][j] = a.paths[i][k] + b.paths[k][j]; // } // } // } // } for (int i = 0; i < C; i++) { for (int j = 0; j < C; j++) { res.paths[i][j] = INF; for (int k = 0; k < C; k++) { res.paths[i][j] = min(res.paths[i][j], a.paths[i][k] + b.paths[k][j]); } } } return res; } }; vector<node> tree; inline int query(int v1, int v2) { return tree[1].paths[v1][v2]; } void update(int n, int tl, int tr, int p) { if (tl == tr) { tree[n] = node(true, tl); return; } int mid = (tl + tr) / 2; if (p <= mid) update(n * 2, tl, mid, p); else update(n * 2 + 1, mid + 1, tr, p); tree[n] = tree[n].merge(tree[n * 2], tree[n * 2 + 1]); } void build(int n, int tl, int tr) { if (tl == tr) { tree[n] = node(true, tl); return; } int mid = (tl + tr) / 2; build(n * 2, tl, mid), build(n * 2 + 1, mid + 1, tr); tree[n] = tree[n].merge(tree[n * 2], tree[n * 2 + 1]); } void init() { tree.clear(); tree.resize(4 * BLOCK_COUNT); build(1, 0, BLOCK_COUNT - 1); } } seg; void init(int R, int C, int h[5000][200], int v[5000][200]) { ::R = R, ::C = C; for (int i = 0; i < R; i++) { for (int j = 0; j < C - 1; j++) { H[i][j] = h[i][j]; } } for (int i = 0; i < R - 1; i++) { for (int j = 0; j < C; j++) { V[i][j] = v[i][j]; } } seg.init(); } void changeH(int P, int Q, int W) { H[P][Q] = W; seg.update(1, 0, BLOCK_COUNT - 1, P / BLOCK_SIZE); } void changeV(int P, int Q, int W) { V[P][Q] = W; seg.update(1, 0, BLOCK_COUNT - 1, P / BLOCK_SIZE); } int escape(int V1, int V2) { return seg.query(V1, V2); }

Compilation message (stderr)

grader.c: In function 'int main()':
grader.c:15:6: warning: variable 'res' set but not used [-Wunused-but-set-variable]
  int res;
      ^~~
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