2019省赛A组第4题 迷宫（BFS+DFS）
题目链接http://oj.ecustacm.cn/problem.php?id=1455
下面是答案提交的代码
#include<iostream>
using namespace std;
int main() {
cout << "DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR";
return 0;
}
用DFS来解题
//其中D、U、L、R 分别表示向下、向上、向左、向右走。\
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
char a[35][55];
char b[4] = {
'D','L','R','U' };
bool vis[35][55] = {
0 };
int fx[4][2] = {
{
1,0},{
0,-1},{
0,1},{
-1,0} };
int mins[35][55];
int p[2000] = {
0 }, q[2000] = {
0 };
int minn=2000, ans=0;//D<L<R<U。
bool fw(int x, int y) {
if (x > 0 && x <=30 && y > 0 && y <= 50)
return true;
return false;
}
void dfs(int x, int y) {
if (ans > minn){
return;
}
if (x == 30&& y == 50) {
if (minn > ans) {
minn = ans;
for (int i = 0; i < ans; i++) {
p[i] = q[i];
}
}
return;
}
vis[x][y] = 1;
for (int i = 0; i < 4; i++) {
int xx = x + fx[i][0], yy = y + fx[i][1];
if (fw(xx, yy) && a[xx][yy] != '1' && !vis[xx][yy]&&ans+1<=mins[xx][yy]) {
q[ans++] = i;
mins[xx][yy] = ans;
dfs(xx, yy);
ans--;
}
}
vis[x][y] = 0;
}
int main() {
for (int i = 1; i <= 30; i++) {
for (int j = 1; j <= 50; j++) {
cin >> a[i][j];
}
}
memset(mins, 0x3f, sizeof(mins));
dfs(1, 1);
cout << minn << endl;
for (int i = 0; i < minn; i++) {
cout << b[p[i]];
}
return 0;
}
用BFS来解题
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstring>
#include<queue>
using namespace std;
typedef struct Node {
int x, y;
string s;
Node(int xx,int yy,string ss):x(xx),y(yy),s(ss){
}
}node;
queue<node> q;
char a[35][55];
char b[4] = {
'D','L','R','U' };
bool vis[35][55] = {
0 };
int fx[4][2] = {
{
1,0},{
0,-1},{
0,1},{
-1,0} };
bool fw(int x, int y) {
if (x > 0 && x <= 30 && y > 0 && y <= 50)
return true;
return false;
}
void bfs() {
node now(1, 1, "");
q.push(now);
while (!q.empty()) {
node temp = q.front();
q.pop();
if (temp.x == 30 && temp.y == 50) {
cout << temp.s;
return;
}
for (int i = 0; i < 4; i++) {
int xx = temp.x + fx[i][0], yy = temp.y + fx[i][1];
if (fw(xx, yy) && a[xx][yy] != '1' && !vis[xx][yy]) {
node neww(xx, yy, temp.s + b[i]);
q.push(neww);
vis[xx][yy] = 1;
}
}
}
}
int main() {
for (int i = 1; i <= 30; i++) {
for (int j = 1; j <= 50; j++) {
cin >> a[i][j];
}
}
vis[1][1] = 1;
bfs();
return 0;
}