炸弹人（广度优先搜索)

   首先用广度优先搜索来枚举出所有小人能够到达的点，并计算出每个点能够消灭的敌人的数量。如果下一个点能够消灭的敌人数量大于前一个，那么更新消灭敌人数量的最大值，并更新坐标点。下面是代码。

#include<stdio.h>#include<queue>#include<string.h>using namespace std;struct A{    int x,y;};int Next[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};int book[20][20],n,m,Max,mx,my;char Map[20][20];void getnum(int i, int j){    int sum=0,x,y;    x = i, y = j;    while(Map[x][y] != '#')    {        if(Map[x][y] == 'G')            sum++;        x--;    }    x = i, y = j;    while(Map[x][y] != '#')    {        if(Map[x][y] == 'G')            sum++;        x++;    }    x = i, y = j;    while(Map[x][y] != '#')    {        if(Map[x][y] == 'G')            sum++;        y--;    }    x = i, y = j;    while(Map[x][y] != '#')    {        if(Map[x][y] == 'G')            sum++;        y++;    }    if(sum > Max)    {        Max = sum;        mx = i;        my = j;    }}void bfs(A head){    A h,t;    int i,tx,ty;    queue<A> q;    getnum(head.x,head.y);    q.push(head);    book[head.x][head.y] = 1;    while(!q.empty())    {        h = q.front();        q.pop();        for(i = 0; i < 4; i++)        {            tx = h.x + Next[i][0];            ty = h.y + Next[i][1];            if(tx < 0 || tx >= n || ty < 0 || ty >= m)                continue;            if(Map[tx][ty] == '.' && book[tx][ty] == 0)            {                t.x = tx;                t.y = ty;                getnum(tx, ty);                q.push(t);                book[tx][ty] = 1;            }            if(tx == n-1 && ty == m-1)                return ;        }    }}int main(){    A h;    int i,j;    while(~scanf("%d%d%d%d",&n,&m,&h.x,&h.y))    {        memset(book,0,sizeof(book));        Max = 0;        mx = h.x;        my = h.y;        for(i = 0; i < n; i++)            scanf("%s",Map[i]);        bfs(h);        printf("(%d, %d) %d\n",mx,my,Max);    }    return 0;}

/*输入样例：
13 13 3 3
#############
#GG.GGG#GGG.#
###.#G#G#G#G#
#.......#..G#
#G#.###.#G#G#
#GG.GGG.#.GG#
#G#.#G#.#.#.#
##G...G.....#
#G#.#G###.#G#
#...G#GGG.GG#
#G#.#G#G#.#G#
#GG.GGG#G.GG#
#############

输出：（7,11）10

*/