ACM_搜索:杭电oj1026:Ignatius and the Princess I

题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1026

题目大意:一个人要从N*M矩形地图的左上角走到右下角.且只能向四个方向走.X表示墙不能走.字符1-9表示怪物并且数字代表怪物的血量.也就是杀死怪物需要该数字的单位时间.人每走一个格子花费1个单位的时间.然后要你按照格式打印最短时间的路径和具体的路径.

简单的最短路径题.直接用bfs做.用二维数组保存路径.由终点往起点递归.就可以把完整的顺序路径加到容器中去.

AC代码:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define Size 200
typedef std::pair<int, int> pairType;

struct Node
{
    int x;
    int y;
    int time;
    //时间优先.
    bool operator<(Node a) const
    {
        return this->time> a.time;
    }
};
int dir[4][2] = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };
char world[Size][Size];
int visit[Size][Size];
Node path[Size][Size];
int m, n;
char cr;
vector filePath;
int num;

void advisitath(int x,int y)
{
    //判断是否到起始点.
    if (x == 0 && y == 0)
    {
        return;
    }
    advisitath(path[x][y].x, path[x][y].y);
    //因为是递归.最好其实是顺序加入路径的.
    filePath.push_back(pairType(path[x][y].x, path[x][y].y));
}
void print()
{
    //把路径加到容器中.
    advisitath(m - 1, n - 1);
    filePath.push_back(pairType(m - 1, n - 1));
    int secNum = 1;

    printf("It takes %d seconds to reach the target position, let me show you the way.\n", num);
    for (int i = 0; i < filePath.size() - 1; ++i)
    {
        printf("%ds:(%d,%d)->(%d,%d)\n", secNum++, filePath[i].first, filePath[i].second, filePath[i + 1].first, filePath[i + 1].second);
        //如果不是正常路.也就是怪物.
        if (world[filePath[i+1].first][filePath[i+1].second] != '.')
        {
            for (int j = 0; j < world[filePath[i+1].first][filePath[i+1].second] - '0'; ++j)
            {
                printf("%ds:FIGHT AT (%d,%d)\n", secNum++, filePath[i+1].first, filePath[i+1].second);
            }
        }
    }
    printf("FINISH\n");
    filePath.clear();
}
int bfs()
{
    //题目要求最短时间.以时间优先即可.
    priority_queue temp;
    Node now, next, s;
    memset(&s, 0, sizeof(s));
    temp.push(s);
    visit[0][0] = 1;

    while (!temp.empty())
    {
        now = temp.top();
        temp.pop();
        //判断是否到达.
        if (now.x == m - 1 && now.y == n - 1)
        {
            return now.time;
        }
        //分别枚举四个方向.
        for (int i = 0; i < 4; ++i)
        {
            next.x = now.x + dir[i][0];
            next.y = now.y + dir[i][1];
            //进行筛选.
            if (next.x >= 0 && next.y >= 0 && next.x < m && next.y < n && visit[next.x][next.y] == 0 && world[next.x][next.y] != 'X')
            {
                visit[next.x][next.y] = 1;
                //记录来的路径.
                path[next.x][next.y].x = now.x;
                path[next.x][next.y].y = now.y;
                if (world[next.x][next.y] == '.')
                {
                //普通路+1.
                    next.time = now.time + 1;
                }
                else
                {
                //遇到怪物.先+1.然后再加怪物的血量.记得读取的是字符.
                    next.time = now.time + world[next.x][next.y] - '0' + 1;
                }
                temp.push(next);
            }
        }
    }
    return -1;
}

int main()
{
    while (scanf("%d%d", &m,&n) != EOF)
    {
        for (int i = 0; i < m; ++i)
        {
        //要先用getchar()把换行符读取掉.
            getchar();
            for (int j = 0; j < n; ++j)
            {
                scanf("%c", &world[i][j]);
                visit[i][j] = 0;
            }
        }
        memset(path, 0, sizeof(path));
        num = bfs();
        if (num == -1)
        {
            printf("God please help our poor hero.\n");
            printf("FINISH\n");
        }
        else
        {
            print();
        }
    }
    return 0;
}