直接将输出的答案复制，cout就好，所以填空题尽量本地DevC++得到答案后，再专门cout吧

貌似填空题一般也没有输入，只有输出 

输入输出

11001000110101000010101100011010011010101011110111
00011011010101001001001010000001000101001110000000
10100000101000100110101010111110011000010000111010
00111000001010100001100010000001000101001100001001
11000110100001110010001001010101010101010001101000
00010000100100000101001010101110100010101010000101
11100100101001001000010000010101010100100100010100
00000010000000101011001111010001100000101010100011
10101010011100001000011000010110011110110100001000
10101010100001101010100101000010100000111011101001
10000000101100010000101100101101001011100000000100
10101001000000010100100001000100000100011110101001
00101001010101101001010100011010101101110000110101
11001010000100001100000010100101000001000111000010
00001000110000110101101000000100101001001000011101
10100101000101000000001110110010110101101010100001
00101000010000110101010000100010001001000100010101
10100001000110010001000010101001010101011111010010
00000100101000000110010100101001000001000000000010
11010000001001110111001001000011101001011011101000
00000110100010001000100000001000011101000000110011
10101000101000100010001111100010101001010000001000
10000010100101001010110000000100101010001011101000
00111100001000010000000110111000000001000000001011
10000001100111010111010001000110111010101101111000
DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR

AC  代码

#include
using namespace std;
int main()
{cout<<"DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR";return 0;
}

2，读文件

🌳BFS二  AC

读入文件只需要增加这些操作

maze.txt要和代码文件在同一目录下

#include
...
//读取文件数据ifstream infile("maze.txt"); //打开maze.txt文件//按行读入maze.txt文件的字符串, 保存到maze数组中for(int i = 0; i < 30; ++i)getline(infile, maze[i]);

👆

原始代码

#include
#include
using namespace std;
int tx, ty, flag, vis[100][100];
string maze[30], ans;
struct node
{   //只要求输出最短路的路线,不要求输出具体步数,不用sint x, y, f; //横坐标, 纵坐标, 父坐标char al; //方向字母
};
int main()
{//读取文件数据ifstream infile("maze.txt"); //打开maze.txt文件//按行读入maze.txt文件的字符串, 保存到maze数组中for(int i = 0; i < 30; ++i)getline(infile, maze[i]);struct node que[1510]; //30 * 50//方向数组, 由于D{1,0},{0,-1},{0,1},{-1,0}};char dir[4] = {'D','L','R','U'}; //方向字母//初始化队列int head = 0, tail = 0;//队列插入迷宫入口坐标que[tail].x = 0;que[tail].y = 0;que[tail].f = 0;que[tail].al = '0';tail++;//队列不为空while(head < tail) {//枚举4个方向for(int i = 0; i < 4; ++i) {//新的坐标tx = que[head].x + next[i][0];ty = que[head].y + next[i][1];//超出范围if(tx < 0 || ty < 0 || tx >= 30 || ty >= 50)continue;//未访问过且不为障碍if(vis[tx][ty] == 0 && maze[tx][ty] == '0') {vis[tx][ty] = 1; //标记que[tail].x = tx;que[tail].y = ty;que[tail].f = head;que[tail].al = dir[i];tail++;//BFS不需要取消标记, 只入队1次}//到达终点if(tx == 29 && ty == 49) {flag = 1;break;}}if(flag) break;head++; //后续点的扩展}//用ans字符串保存从尾到头输出的路径tail--; //最后的tail是队尾下一个,所以-1while(tail) {ans += que[tail].al; //字符的连接tail = que[tail].f; //更新队尾为上一个点}for(int i = ans.size() - 1; i >= 0; --i)cout<

#include
using namespace std;
int main()
{cout<<"DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR";return 0;
}

#include
using namespace std;
int main()
{cout<<490;return 0;
}

#include
using namespace std;
int book[30], ans; //book[]的大小要大于20
int c[20][5] = {
{97, 90, 0 ,0, 0},
{92, 85, 96, 0, 0},
{0, 0, 0, 0, 93},
{0, 0, 0, 80, 86},
{89, 83, 97, 0, 0},
{82, 86, 0, 0, 0},
{0, 0, 0, 87, 90},
{0, 97, 96, 0, 0},
{0, 0, 89, 0, 0},
{95, 99, 0, 0, 0},
{0, 0, 96, 97, 0},
{0, 0, 0, 93, 98},
{94, 91, 0, 0, 0},
{0, 83, 87, 0, 0},
{0, 0, 98, 97, 98},
{0, 0, 0, 93, 86},
{98, 83, 99, 98, 81},
{93, 87, 92, 96, 98},
{0, 0, 0, 89, 92},
{0, 99, 96, 95, 81}
};
void dfs(int step, int now) //当前第几号位
{if(step == 6) { //完成5个位置后, 结束递归ans = max(ans, now);return; //递归回上一个位置}//遍历for(int i = 0; i < 20; ++i) {if(book[i] == 0) { //第i号队员没被选上book[i] = 1; //标记dfs(step + 1, now + c[i][step - 1]); //递归book[i] = 0; //取消标记}}
}
int main()
{dfs(1, 0); //从1号位开始cout<

#include //ifstream读取文件
...
//ifstream打开文件,命名为infileifstream infile("team.txt"); //文件名作为参数,""括起来//读取文件每一个数字for(int i = 0; i < 20; ++i)for(int j = 0; j < 6; ++j)infile >> c[i][j]; //存储到二维数组cinfile.close(); //关闭文件

#include
#include //ifstream读取文件
using namespace std;
int book[30], ans, c[20][6];
void dfs(int step, int now) //当前第几号位
{if(step == 6) { //完成5个位置后, 结束递归ans = max(ans, now);return; //递归回上一个位置}//遍历for(int i = 0; i < 20; ++i) {if(book[i] == 0) { //第i号队员没被选上book[i] = 1; //标记dfs(step + 1, now + c[i][step]); //递归book[i] = 0; //取消标记}}
}
int main()
{//ifstream打开文件,命名为infileifstream infile("team.txt"); //文件名作为参数,""括起来//读取文件每一个数字for(int i = 0; i < 20; ++i)for(int j = 0; j < 6; ++j)infile >> c[i][j]; //存储到二维数组cinfile.close(); //关闭文件dfs(1, 0); //从1号位开始cout<

#include
using namespace std;
int main()
{cout<<490;return 0;
}

#include
...
//读取文件数据ifstream infile("maze.txt"); //打开maze.txt文件//按行读入maze.txt文件的字符串, 保存到maze数组中for(int i = 0; i < 30; ++i)getline(infile, maze[i]);