usaco2.4.2 Overfencing
来源:互联网 发布:吉林大学就业知乎 编辑:程序博客网 时间:2024/06/11 20:56
一 原题
Kolstad and Schrijvers
Farmer John went crazy and created a huge maze of fences out in a field. Happily, he left out two fence segments on the edges, and thus created two "exits" for the maze. Even more happily, the maze he created by this overfencing experience is a `perfect' maze: you can find a way out of the maze from any point inside it.
Given W (1 <= W <= 38), the width of the maze; H (1 <= H <= 100), the height of the maze; 2*H+1 lines with width 2*W+1 characters that represent the maze in a format like that shown later - then calculate the number of steps required to exit the maze from the `worst' point in the maze (the point that is `farther' from either exit even when walking optimally to the closest exit). Of course, cows walk only parallel or perpendicular to the x-y axes; they do not walk on a diagonal. Each move to a new square counts as a single unit of distance (including the move "out" of the maze.
Here's what one particular W=5, H=3 maze looks like:
+-+-+-+-+-+| |+-+ +-+ + +| | | |+ +-+-+ + +| | | +-+ +-+-+-+
Fenceposts appear only in odd numbered rows and and odd numbered columns (as in the example). The format should be obvious and self explanatory. Each maze has exactly two blank walls on the outside for exiting.
PROGRAM NAME: maze1
INPUT FORMAT
Line 1:W and H, space separatedLines 2 through 2*H+2:2*W+1 characters that represent the mazeSAMPLE INPUT (file maze1.in)
5 3+-+-+-+-+-+| |+-+ +-+ + +| | | |+ +-+-+ + +| | | +-+ +-+-+-+
OUTPUT FORMAT
A single integer on a single output line. The integer specifies the minimal number of steps that guarantee a cow can exit the maze from any possible point inside the maze.SAMPLE OUTPUT (file maze1.out)
9The lower left-hand corner is *nine* steps from the closest exit.
二 分析
三 代码
/*ID:maxkibb3LANG:C++PROG:maze1*/#include<cstdio>#include<vector>#include<queue>#include<cstring>#include<algorithm>using namespace std;const int MAX_H = 100;const int MAX_W = 38;struct Point { int x, y;};Point Pt(int _x, int _y) { Point ret; ret.x = _x, ret.y = _y; return ret;}int w, h, ans;char map[2 * MAX_H + 1][2 * MAX_W + 1];vector<Point> exitp;int v[2 * MAX_H + 1][2 * MAX_W + 1];int v_copy[2 * MAX_H + 1][2 * MAX_W + 1];int dx[4][2] = {{-2, 0}, {2, 0}, {0, -2}, {0, 2}};void check_exit(Point p) { if(map[p.x][p.y] == ' ') { if(p.x == 0) p.x = -1; else if(p.x == h - 1) p.x = h; if(p.y == 0) p.y = -1; else if(p.y == w - 1) p.y = w; exitp.push_back(p); }}void init() { scanf("%d%d", &w, &h); w = 2 * w + 1; h = 2 * h + 1; for(int i = 0; i < h; i++) { getchar(); for(int j = 0; j < w; j++) scanf("%c", &map[i][j]); }}bool movable(Point p, int i) { if(p.x + dx[i][0] < 1 || p.x + dx[i][0] > h - 2 || p.y + dx[i][1] < 1 || p.y + dx[i][1] > w - 2) { return false; } if(map[p.x + dx[i][0] / 2][p.y + dx[i][1] / 2] != ' ') { return false; } else return true;}void bfs(Point s) { memset(v, 0, sizeof(v)); queue<Point> q; q.push(s); v[s.x][s.y] = 0; while(!q.empty()) { Point p = q.front(); q.pop(); for(int i = 0; i < 4; i++) { if(!movable(p, i)) continue; int nx = p.x + dx[i][0], ny = p.y + dx[i][1]; if(v[nx][ny] != 0) continue; q.push(Pt(nx, ny)); v[nx][ny] = v[p.x][p.y] + 1; } }}int main() { freopen("maze1.in", "r", stdin); freopen("maze1.out", "w", stdout); init(); for(int i = 1; i < w; i += 2) { check_exit(Pt(0, i)); check_exit(Pt(h - 1, i)); } for(int i = 1; i < h; i += 2) { check_exit(Pt(i, 0)); check_exit(Pt(i, w - 1)); } bfs(exitp[0]); memcpy(v_copy, v, sizeof(v)); bfs(exitp[1]); for(int i = 1; i < h; i += 2) { for(int j = 1; j < w; j += 2) { int val = min(v[i][j], v_copy[i][j]); ans = max(ans, val); } } printf("%d\n", ans); return 0;}
- USACO2.4.2 Overfencing
- USACO2.4.2 Overfencing (maze1)
- USACO2.4.2 Overfencing (maze1)
- usaco2.4.2 Overfencing
- [USACO2.4]穿越栅栏 Overfencing
- Overfencing
- usaco2.4.2两只塔姆沃斯牛
- 【USACO2.4.2】穿越栅栏
- USACO2.4.2 穿越栅栏
- usaco2
- usaco2
- usaco-Overfencing
- usaco overfencing
- [U]Overfencing
- usaco Overfencing
- USACO:Overfencing
- USACO-Overfencing
- USACO Overfencing
- Objective-C 编码规范
- 介绍几个图处理和流处理大数据平台
- 【LeetCode】467. Unique Substrings in Wraparound String
- ionic—cordova如何一次上传多张图片(类似input file提交表单)
- zzulioj 1963:Deadline
- usaco2.4.2 Overfencing
- 文件按层级打印
- IO流总结
- 牛顿迭代法
- 蓝桥杯第六届 无穷分数(递归)
- ionic+angular 真机调试
- Solution of '1045 - access denied for user 'root'@'localhost'(using password: YES)' in mysql
- 初学SVN配置以及简单使用
- Alisql源码编译安装(详细篇)