目錄

• F. Honeycomb (2018-ACM/ICPC焦作)

F. Honeycomb (2018-ACM/ICPC焦作)

Problem F. Honeycomb
Input file: standard input
Output file: standard output

A honeycomb is a mass wax cells built by honey bees, which can be described as a regular tiling of the
Euclidean plane, in which three hexagons meet at each internal vertex. The internal angle of a hexagon is
120 degrees, so three hexagons at a point make a full 360 degrees. The following figure shows a complete
honeycomb with 3 rows and 4 columns.

Here we guarantee that the first cell in the second column always locates in the bottom right side of
the first cell in the first column, as shown above. A general honeycomb may, on the basis of a complete
honeycomb, lose some walls between adjacent cells, but the honeycomb is still in a closed form. A possible
case looks like the figure below.

Hamilton is a brave bee living in a general honeycomb. Now he wants to move from a starting point to
a specified destination. The image below gives a feasible path in a 3×4 honeycomb from the 1-st cell in
the 2-nd column to the 1-st cell in the 4-th column.

Please help him find the minimum number of cells that a feasible path has to pass through (including
the starting point and the destination) from the specified starting point to the destination.

Input
The input contains several test cases, and the first line contains a positive integer T indicating the number
of test cases which is up to 1e4.
For each test case, the first line contains two integers r and c indicating the number of rows and the
number of columns of the honeycomb, where 2 <= r, c <= 1e3.
The following (4r+3) lines describe the whole given honeycomb, where each line contains at most (6c+3)
characters. Odd lines contain grid vertices represented as plus signs (“+”) and zero or more horizontal
edges, while even lines contain two or more diagonal edges. Specifically, a cell is described as 6 vertices
and at most 6 edges. Its upper boundary or lower boundary is represented as three consecutive minus
signs (“-”). Each one of its diagonal edges, if exists, is a single forward slash (“/”) or a single backslash
(“”) character. All edge characters will be placed exactly between the corresponding vertices. At the
center of the starting cell (resp. the destination), a capital “S” (resp. a capital “T”) as a special character
is used to indicate the special cell. All other characters will be space characters. Note that if any input
line could contain trailing whitespace, that whitespace will be omitted.
We guarantee that all outermost wall exist so that the given honeycomb is closed, and exactly one “S”
and one “T” appear in the given honeycomb. Besides, the sum of r · c in all test cases is up to 2e6.

Output
For each test case, output a line containing the minimum number of cells that Hamilton has to visit moving
from the starting cell (“S”) to the destination (“T”), including the starting cell and the destination. If
no feasible path exists, output -1 instead.

題意是二維平面最短路，不過地圖形成蜂窩一樣的。我們需要做一個映射。這題題目內存給了一個G，%>_<%，！！！知道這一點后就很好做了。現場題目找不到內存時間，網頁提交的地方也沒寫，下載輸入輸出的地方才寫了。賽后才找到。唉。。。

具體看代碼把，映射點

用map<pair<int, int>, vector<pair<int, int> > > E;存圖，然后bfs。

代碼在學校windows都跑不起來，內存用太多了。

注意memset復雜度可能不太對。因為T比較多。

標記點不要用set<pair<int, int> >s了，直接開一個二維數組。內存不要命的開

/** 1 3 412345678901234567890123456789 1 +---+ | +---+ | 2 / | \ | / | \ | 3+---O +---+ O +---+ [3, 5] [3, 17] 4 \ | | \ / | \ 5 + + S +---+ T + [5, 11] [5, 23] 6 / \ / / 7+ +---+ + + [7, 5] [7, 17] 8 \ \ / \ 9 +---+ +---+ + [9, 11] [9, 23] 0 / / 1+ +---+ + + 2 \ / \ 3 +---+ +---+ + 4 \ / \ / 5 +---+ +---+*/ #include <bits/stdc++.h>using namespace std; const int MAX_R = (1e3 + 10) * 4 + 3; const int MAX_C = (1e3 + 10) * 6 + 3;char mp[MAX_R][MAX_C];bool point_flag[MAX_R][MAX_C];int t, r, c;int Next[6][2] = {-1, -3, -1, 3, 1, -3, 1, 3, -2, 0, 2, 0};//set<pair<int, int> > point_flag;map<pair<int, int>, vector<pair<int, int> > > E;void init() {for(int i = 3; i <= MAX_R; i += 4) {for(int j = 5; j <= MAX_C; j += 12) {//point_flag.insert(make_pair(i, j));point_flag[i][j] = true;}}for(int i = 5; i <= MAX_R; i += 4) {for(int j = 11; j <= MAX_C; j += 12) {//point_flag.insert(make_pair(i, j));point_flag[i][j] = true;}} }bool isin(int x, int y) {return x >= 1 && x <= 4 * r + 3 && y >= 1 && y <= 6 * c + 3; }struct Node {pair<int, int>pos;int step;Node() {}Node(int xx, int yy, int st) {step = st;pos = make_pair(xx, yy);}Node(pair<int, int>p, int st) {step = st;pos = p;} };int bfs(pair<int, int> &begin_s, pair<int, int> &end_t) {queue<Node>que;set<pair<int, int> >vis;que.push(Node(begin_s, 1));vis.insert(begin_s);while(!que.empty()) {Node now = que.front();que.pop();if(now.pos == end_t) {return now.step;}for(auto it: E[now.pos]) {if(!vis.count(it)) {vis.insert(it);mp[it.first][it.second] = '#';que.push(Node(it, now.step + 1));}}}return -1; }int main() {init();scanf("%d", &t);while(t--) {E.clear();scanf("%d %d", &r, &c);getchar();for(int i = 1; i <= 4 * r + 3; i++ ) {fgets(mp[i] + 1, MAX_C, stdin);}pair<int, int>begin_s, end_t;for(int i = 1; i <= 4 * r + 3; i++ ) {for(int j = 1; mp[i][j] != '\n'; j++ ) {if(point_flag[i][j]) {if(mp[i][j] == 'S') {begin_s = make_pair(i, j);}if(mp[i][j] == 'T') {end_t = make_pair(i, j);}for(int k = 0; k < 6; k++ ) {int gx = i + Next[k][0];int gy = j + Next[k][1];int mx = i + Next[k][0] * 2;int my = j + Next[k][1] * 2;if(!isin(gx, gy) || !isin(mx, my) || mp[gx][gy] != ' ') {continue;}if(mp[mx][my] == 'S' || mp[mx][my] == 'T' || mp[mx][my] == ' ') {E[make_pair(i, j)].push_back(make_pair(mx, my));}}}}}printf("%d\n", bfs(begin_s, end_t));}return 0; }

轉載于:https://www.cnblogs.com/Q1143316492/p/10092877.html

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