搜索

1.对1、2、3 ，3个数全排列（排列考虑顺序）

//万能的搜索－递归
        dfs(1);

递归

int a[3],book[3],n=3;

void dfs(int step)
{
    //如果站在n+1个盒子面前，则表示前n个盒子已经放好的牌
    if (step == n+1)
    {
        for (int i = 1; i <= n; i++)
        {
            printf("%d",a[i]);
        }
        printf("
");
        return;//返回之前的一步（最近一次调用dfs函数的地方）
    }
    
    //step表示第step个盒子
    for (int i = 1; i <= n; i++)
    {
        if (book[i] == 0)
        {
            a[step] = i;
            book[i] = 1;//该牌已经打出
            dfs(step+1);
            book[i] = 0;//收回该牌
        }
    }
}

 2.求到达目的地的最小步数

 广度优先搜索(队列＋hashMap之类的去重集合)和深度优先搜索（递归）

int q = 3,p = 2,min = 1000;
int book[51][51];

int myMap[51][51] = {{0,0,1,0},
                    {0,0,0,0},
                    {0,0,1,0},
                    {0,1,0,0},
                    {0,0,0,1}};

int next[4][2] = {{0,1},
                    {1,0},
                    {0,-1},
                    {-1,0}};


void dfs(int x,int y,int step);


struct note{
    int x;//横坐标
    int y;//纵坐标
    int s;//步数
    int f;//父亲在队列中的编号，本题不要求输出路径，可以不需要f
};


int main(int argc, const char * argv[]) {
    @autoreleasepool {
        //1深度优先搜索
//        dfs(0, 0, 0);
//        printf("min = %d
",min);
        //2广度优先搜索
        struct note que[2501];
        int head = 0,tail = 0;
        int tx,ty,flag = 0;
        que[tail].x = 0;
        que[tail].y = 0;
        que[tail].f = 0;
        que[tail].s = 0;
        tail++;
        book[0][0] = 1;
        while (head < tail)
        {
            for (int k = 0; k < 4; k++)
            {
                tx = que[head].x + next[k][0];
                ty = que[head].y + next[k][1];
                if (tx < 0 || ty < 0 || tx > 5 || ty > 4) {
                    continue;
                }
                if (myMap[tx][ty]==0 && book[tx][ty]==0) {
                    book[tx][ty] = 1;
                    //插入新的点到队列中
                    que[tail].x = tx;
                    que[tail].y = ty;
                    que[tail].f = head;
                    que[tail].s = que[head].s + 1;
                    tail++;
                }
                //到达目的地
                if (tx == q && ty == p)
                {
                    flag = 1;
                    break;
                }
            }
            if (flag==1) {
                break;
            }
            head++;
        }
        printf("dfd = %d
",que[tail-1].s);
        
    }
    return 0;
}

void dfs(int x,int y,int step)
{
    int tx,ty;
　　//判断是否到达结束位置，结束递归
    if (x == q && y == p)
    {
        if (step < min) {
            min = step;
        }
        return;
    }
    //遍历4个方向
    for (int k = 0; k < 4; k++) {
　　　　//计算下一个点的坐标tx、ty
        tx = x + next[k][0];
        ty = y + next[k][1];
        //判断是否越界
        if (tx < 0 || ty < 0 || ty > 5 || tx > 4)
        {
            continue;
        }
        //判断该点是否障碍物和已经走过
        if (myMap[tx][ty] == 0 && book[tx][ty]==0)
        {
            book[tx][ty] = 1;
            dfs(tx, ty, step+1);
            book[tx][ty] = 0;
            
        }
    }
}

3.炸弹人（求消灭敌人的最大个数）

3.1广度优先搜索实现

int next[4][2] = {{0,1},
                    {1,0},
                    {0,-1},
                    {-1,0}};


int a[21][21] = {
{'#','#','#','#','#','#','#','#','#','#','#','#','#'},
{'#','G','G','.','G','G','G','#','G','G','G','.','#'},
{'#','#','#','.','#','G','#','G','#','G','#','G','#'},
{'#','.','.','.','.','.','.','.','#','.','.','G','#'},
{'#','G','#','.','#','#','#','.','#','G','#','G','#'},
{'#','G','G','.','G','G','G','.','#','.','G','G','#'},
{'#','#','G','.','.','G','#','.','.','.','.','.','#'},
{'#','G','#','.','#','G','#','#','#','.','#','G','#'},
{'#','.','.','.','#','#','#','#','#','#','#','#','#'},
{'#','G','#','.','#','G','#','G','#','.','#','G','#'},
{'#','G','G','.','G','G','G','#','G','.','G','G','#'},
{'#','#','#','#','#','#','#','#','#','#','#','#','#'}};




struct note{
    int x;//横坐标
    int y;//纵坐标
    int s;//步数
    int f;//父亲在队列中的编号，本题不要求输出路径，可以不需要f
};

int getnum(int i,int j);

.

//3广度优先搜索。用炸弹消灭最多的敌人G、.代表可走的路、＃代表墙
        struct note que[401];
        int head = 0,tail = 0;
        int book[20][20] = {0};
        int sum,max,mx,my,tx,ty;
        que[tail].x = 3;
        que[tail].y = 3;
        tail++;
        book[3][3] = 1;
        max = getnum(3, 3);
        mx = 3;
        my = 3;
        while (head<tail) {
            for (int k = 0;k < 4; k++)
            {
                tx = que[head].x + next[k][0];
                ty = que[head].y + next[k][1];
                
                //判断是否越界
                if (tx < 0 || ty < 0 || ty > 13 || tx > 12)
                {
                    continue;
                }
                //
                if (a[tx][ty]=='.' && book[tx][ty]==0) {
                    book[tx][ty] = 1;
                    que[tail].x = tx;
                    que[tail].y = ty;
                    tail++;
                    
                    sum = getnum(tx, ty);
                    printf("____h sum = %zd
",sum);
                    if (sum>max) {
                        max = sum;
                        mx = tx;
                        my = ty;
                    }
                }
            }
            head++;
        }
        printf("haha = %d,%d ,count = %d",mx,my,max);

.

int getnum(int i,int j)
{
    int sum = 0,x,y;
    x = i;y = j;
    while (a[x][y]!='#') {
        if (a[x][y]=='G') {
            sum++;
        }
        x--;
    }
    x = i;y = j;
    while (a[x][y]!='#') {
        if (a[x][y]=='G') {
            sum++;
        }
        x++;
    }
    x = i;y = j;
    while (a[x][y]!='#') {
        if (a[x][y]=='G') {
            sum++;
        }
        y--;
    }
    x = i;y = j;
    while (a[x][y]!='#') {
        if (a[x][y]=='G') {
            sum++;
        }
        y++;
    }
    return sum;
}

 3.2深度优先搜索实现

//深度优先搜索
        book[3][3] = 1;
        max = getnum(3, 3);
        dfs(3, 3, 0);
        printf("haha = %d,%d ,count = %d",mx,my,max);

.

void dfs(int x,int y,int step)
{
    int sum = getnum(x, y);
    if (sum > max) {
        max = sum;
        mx = x;
        my = y;
    }
    int tx,ty;
    for (int k = 0; k < 4; k++) {
        tx = x + next[k][0];
        ty = y + next[k][1];
        if (tx < 0 || ty < 0 || ty > 13 || tx > 12) {
            continue;
        }
        if (a[tx][ty]=='.' && book[tx][ty]==0) {
            book[tx][ty] = 1;
            dfs(tx, ty, step+1);
        }
    }
}