hdu 1026 Ignatius and the Princess I bfs

用到了图论最短路更新的思想，当走到走过的地方时要更新

数据：

5 6

.XX.1.

..X.6.

2...X.

...XX.

XXXXX.

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

struct node{
    int x,y,num,t;
}a;

struct Path{
    int x,y,pre,fight;
}p[500000];

int n,m,k,l,key,dp[105][105];
char map[105][105];

void f(int z)
{
    int i;
    if(z)
    {
        f(p[z].pre);
        printf("%ds:(%d,%d)->(%d,%d)\n",l++,p[p[z].pre].x,p[p[z].pre].y,p[z].x,p[z].y);
        for (i=0;i<p[z].fight;i++)
            printf("%ds:FIGHT AT (%d,%d)\n",l++,p[z].x,p[z].y);
    }
}

void bfs()
{
    int i,row,col,time,dir[4][2]={0,1,1,0,-1,0,0,-1};
    queue<node> q;
    q.push(a);
    while (!q.empty())
    {
        a=q.front();q.pop();
        for (i=0;i<4;i++)
        {
            row=dir[i][0]+a.x;
            col=dir[i][1]+a.y;
            if(row>=0&&col>=0&&row<n&&col<m&&map[row][col]!='X')
            {
                if(map[row][col]=='.')time=1;
                else time=map[row][col]-'0'+1;
                if(dp[row][col]>a.t+time)
                {
                    if(row==n-1&&col==m-1)key=k;
                    dp[row][col]=a.t+time;
                    p[k].pre=a.num;
                    p[k].x=row;
                    p[k].y=col;
                    p[k].fight=time-1;
                    node b;
                    b.x=row;
                    b.y=col;
                    b.num=k++;
                    b.t=a.t+time;
                    q.push(b);
                }
            }
        }
    }
}

int main()
{
    int i,j;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        for (i=0;i<n;i++)
        {
            for (j=0;j<m;j++)
                dp[i][j]=0x3f3f3f3f;
        }
        dp[0][0]=0;
        for (i=0;i<n;i++)scanf("%s",map[i]);
        a.x=a.y=a.num=a.t=0;
        k=l=1;
        memset(p,0,sizeof(p));
        bfs();
        if(dp[n-1][m-1]==0x3f3f3f3f)printf("God please help our poor hero.\n");
        else
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.\n",dp[n-1][m-1]);
            f(key);
        }
        printf("FINISH\n");
    }
    return 0;
}