XiangTan Univ. invitational Contest 2010 I,ROBOT
来源:互联网 发布:网络作家富豪榜发布 编辑:程序博客网 时间:2024/06/06 04:06
广搜,之前用四个方向不标记的广搜方法错了。。记得标记四个方向,这题很纠结。。。
# include<iostream>
#include <queue>
using namespace std;
struct Node
{
int x, y;
int dir;
int step;
int judge[4];
bool operator <(const Node t)const
{
return step > t.step;
}
}start;
Node next, pre;
const int Go[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int n, m, ex, ey;
char map[110][110];
bool visitedm[110][110][4];
int ans;
bool check(int x, int y)
{
return (x >= 0&& x < n && y >= 0 && y < m);
}
void Bfs()
{
priority_queue <Node> Que;
memset(visitedm, false, sizeof(visitedm));
start.judge[0] = 1;
Que.push(start);
int kkk = 0;
while (!Que.empty()){
pre = Que.top();
Que.pop();
kkk ++;
if (map[pre.x][pre.y] == 'T'){
ans = pre.step;
return ;
}
for (int i = 0; i < 4; i ++){
if (i == pre.dir){
int xx = pre.x + Go[i][0];
int yy = pre.y + Go[i][1];
if (check(xx, yy) && !visitedm[xx][yy][i] && map[xx][yy] != '#'){
visitedm[xx][yy][i] = true;
next.step = pre.step+1;
if (map[xx][yy] == 'T'){
ans = next.step;
return ;
}
next.x = xx, next.y = yy;
next.dir = i;
Que.push(next);
}
}
else {
if (pre.dir == 0 && i == 1)continue;
if (pre.dir == 1 && i == 0)continue;
if (pre.dir == 2 && i == 3)continue;
if (pre.dir == 3 && i == 2)continue;
next.x = pre.x, next.y = pre.y;
next.dir = i;
next.step = pre.step + 1;
if (!visitedm[pre.x][pre.y][i]){
visitedm[pre.x][pre.y][i] = true;
Que.push(next);
}
}
}
}
}
int main()
{
int t;
scanf("%d", &t);
while (t --){
scanf("%d %d", &n, &m);
for (int i = 0; i <n; i ++){
scanf("%s", map[i]);
for (int j = 0; j < m; j ++){
if (map[i][j] == 'S'){
start.x = i;
start.y = j;
start.step = 0;
start.dir = 0;
memset(start.judge, 0, sizeof(start.judge));
}
}
}
ans = -1;
Bfs();
printf("%d/n", ans);
}
return 0;
}
- XiangTan Univ. invitational Contest 2010 I,ROBOT
- XiangTan Univ. invitational Contest 2010Allocation of Memory
- Xiangtan Invitation Contest 2017
- Xiangtan Invitation Contest 2017
- 2014 ACM-ICPC Beijing Invitational Programming Contest
- Xiangtan Invitation Contest 2017(2017湘潭邀请赛)(Done)
- The University of Chicago Invitational Programming Contest 2012 解题报告
- 2014 ACM-ICPC Beijing Invitational Programming Contest E
- hdu3853 LOOPS 2011 Invitational Contest Host by BUPT
- TopCoder 2001 Invitational Semifinals A - Division I, Level One
- ACM ICPC 2017 Warmup Contest 6(ACM Google Cup 2011 Invitational Programming Contest)
- BNU 34990 Justice String 2014 ACM-ICPC Beijing Invitational Programming Contest
- hdu 1698 Just a Hook || 2008 “Sunline Cup” National Invitational Contest || 线段树
- HDU 6058 Kanade's sum (链表, 2017 Multi-Univ Training Contest 3)
- HDU 6059 Kanade's trio (字典树, 2017 Multi-Univ Training Contest 3)
- HDU 6065 RXD, tree and sequence (LCA, 2017 Multi-Univ Training Contest 3)
- HDU 6071 Lazy Running (Dijstra, 2017 Multi-Univ Training Contest 4)
- HDU 6078 Wavel Sequence (dp + 树状数组, 2017 Multi-Univ Training Contest 4)
- C#/CSharp - 禁用Form的关闭按钮
- 解决PS1的显示问题,busybox的/u /h特殊字符解析问题
- *(字符型 *)addr=data;
- DSP浮点数研究之一:小数的二进制表示
- expdp/impdp 表空间模式迁移实施实验
- XiangTan Univ. invitational Contest 2010 I,ROBOT
- ASP.NET页面传值的方法 和一些实用技巧
- 前导声明代替头文件依赖和交叉包含
- IIS网站属性中没有ASP.NET选项
- 银行110自动监控报警系统解决方案
- BaiDu 一面 之后
- 利用TI的fftlib进行简单的FFT
- hadoop 作业提交
- NDK Shell for Linux/Cygwin(Windows)