poj 2049

Description

Nemo is a naughty boy. One day he went into the deep sea all by himself. Unfortunately, he became lost and couldn‘t find his way home. Therefore, he sent a signal to his father, Marlin, to ask for help. 
After checking the map, Marlin found that the sea is like a labyrinth with walls and doors. All the walls are parallel to the X-axis or to the Y-axis. The thickness of the walls are assumed to be zero. 
All the doors are opened on the walls and have a length of 1. Marlin cannot go through a wall unless there is a door on the wall. Because going through a door is dangerous (there may be some virulent medusas near the doors), Marlin wants to go through as few doors as he could to find Nemo. 
Figure-1 shows an example of the labyrinth and the path Marlin went through to find Nemo. 

We assume Marlin‘s initial position is at (0, 0). Given the position of Nemo and the configuration of walls and doors, please write a program to calculate the minimum number of doors Marlin has to go through in order to reach Nemo.

Input

Output

Sample Input

8 9
1 1 1 3
2 1 1 3
3 1 1 3
4 1 1 3
1 1 0 3
1 2 0 3
1 3 0 3
1 4 0 3
2 1 1
2 2 1
2 3 1
3 1 1
3 2 1
3 3 1
1 2 0
3 3 0
4 3 1
1.5 1.5
4 0
1 1 0 1
1 1 1 1
2 1 1 1
1 2 0 1
1.5 1.7

1 6
1 2 1 5
0 2 0
0 3 0
0 4 0
0 5 0
0 6 0
0 7 0
0.5 5.5

4 1
3 2 1 1
1 3 0 2
1 1 0 2
1 1 1 2
2 1 1
1.5 1.5
-1 -1

Sample Output

5
-1

2

0

题目意思很清晰了，直接上AC代码了，里面解释很清楚

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
#define MM 210
#define MX 9999999
struct Node{
    int x,y,dis;
    friend bool operator <(const Node &a,const Node &b){
        return a.dis>b.dis;                                 //注意这里是>,因为优先队列是大顶堆结构 
    }
    Node(int X,int Y,int DIS){
        x=X; y=Y; dis=DIS;
    }
};
int xe[MM][MM],ye[MM][MM];             //xe表示横向边  ye表示纵向边 
int dis[MM][MM];                       //表示到 0 0的距离 
int f_x,f_y;
priority_queue <Node> q;               //优先队列 
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int M,N;
    while(scanf("%d%d",&M,&N)!=EOF&&(M!=-1 || N!=-1)){
        memset(xe,0,sizeof(xe));
        memset(ye,0,sizeof(ye));
        int x,y,d,t;
        int mx,my;
        mx=my=0;
        for(int i=0;i<M;i++){                    //初始化墙 
            scanf("%d%d%d%d",&x,&y,&d,&t);
            if(d==1){
                for(int j=0;j<t;j++)
                    ye[x][y+j]=MX;               //ye[x][y]表示 x y+1这条边 
                mx=max(mx,x);
                my=max(my,y+t);
            }
            else{
                for(int j=0;j<t;j++)
                    xe[x+j][y]=MX;
                mx=max(mx,x);
                my=max(my,y+t);
            }
        }
        
        for(int i=0;i<N;i++){                     //初始化门 
            scanf("%d%d%d",&x,&y,&d);
            if(d==1){
                ye[x][y]=1;
                mx=max(mx,x);
                my=max(my,y+1);
            }
            else{
                xe[x][y]=1;
                mx=max(mx,x+1);
                my=max(my,y);
            }
        }
        float xd,yd; scanf("%f%f",&xd,&yd);
        if(xd>mx || yd>my){
            printf("0\n");
            continue;
        }
        
        f_x=(int)xd; f_y=(int)yd;               
        for(int i=0;i<=mx;i++)
            for(int j=0;j<=my;j++)
                dis[i][j]=MX;
        dis[0][0]=0;
        int tmp_x,tmp_y;
        int flag=0;
        while(!q.empty())
            q.pop();
        q.push(Node(0,0,0));
        while(!q.empty()){                         //从0 0出发bfs直到到达目的地 
            tmp_x=q.top().x;  tmp_y=q.top().y;
            q.pop();
            if(tmp_x==f_x && tmp_y==f_y){          //到达目的地,退出 
                flag=1;
                break;
            }
            if(tmp_y+1<=my && dis[tmp_x][tmp_y+1]>dis[tmp_x][tmp_y]+xe[tmp_x][tmp_y+1]){   //向上走 
                dis[tmp_x][tmp_y+1]=dis[tmp_x][tmp_y]+xe[tmp_x][tmp_y+1];
                q.push(Node(tmp_x,tmp_y+1,dis[tmp_x][tmp_y+1]));
            }
            if(tmp_x+1<=mx && dis[tmp_x+1][tmp_y]>dis[tmp_x][tmp_y]+ye[tmp_x+1][tmp_y]){    //向右走 
                dis[tmp_x+1][tmp_y]=dis[tmp_x][tmp_y]+ye[tmp_x+1][tmp_y];
                q.push(Node(tmp_x+1,tmp_y,dis[tmp_x+1][tmp_y]));
            }
            if(tmp_y-1>=0 && dis[tmp_x][tmp_y-1]>dis[tmp_x][tmp_y]+xe[tmp_x][tmp_y]){        //向下走  注意：好好体会这里的 +xe[tmp_x][tmp_y] 
                dis[tmp_x][tmp_y-1]=dis[tmp_x][tmp_y]+xe[tmp_x][tmp_y];                                      //而不是 +xe[tmp_x][tmp_y-1]
                q.push(Node(tmp_x,tmp_y-1,dis[tmp_x][tmp_y-1]));
            }
            if(tmp_x-1>=0 && dis[tmp_x-1][tmp_y]>dis[tmp_x][tmp_y]+ye[tmp_x][tmp_y]){        //向左走  注意：好好体会这里的 +ye[tmp_x][tmp_y] 
                dis[tmp_x-1][tmp_y]=dis[tmp_x][tmp_y]+ye[tmp_x][tmp_y];                                      //而不是 +xe[tmp_x][tmp_y-1]
                q.push(Node(tmp_x-1,tmp_y,dis[tmp_x-1][tmp_y]));
            }
        }
        if(flag)
            printf("%d\n",dis[f_x][f_y]);
        else
            printf("-1\n");
    }
    return 0;
}

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poj 2049

原文：http://blog.csdn.net/my_acm/article/details/27565081