poj 2049 Finding Nemo

Finding Nemo

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 8837 Accepted: 2080

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

The input consists of several test cases. Each test case is started by two non-negative integers M and N. M represents the number of walls in the labyrinth and N represents the number of doors. 
Then follow M lines, each containing four integers that describe a wall in the following format: 
x y d t 
(x, y) indicates the lower-left point of the wall, d is the direction of the wall -- 0 means it's parallel to the X-axis and 1 means that it's parallel to the Y-axis, and t gives the length of the wall. 
The coordinates of two ends of any wall will be in the range of [1,199]. 
Then there are N lines that give the description of the doors: 
x y d 
x, y, d have the same meaning as the walls. As the doors have fixed length of 1, t is omitted. 
The last line of each case contains two positive float numbers: 
f1 f2 
(f1, f2) gives the position of Nemo. And it will not lie within any wall or door. 
A test case of M = -1 and N = -1 indicates the end of input, and should not be processed.

Output

For each test case, in a separate line, please output the minimum number of doors Marlin has to go through in order to rescue his son. If he can't reach Nemo, output -1.

Sample Input

8 91 1 1 32 1 1 33 1 1 34 1 1 31 1 0 31 2 0 31 3 0 31 4 0 32 1 12 2 12 3 13 1 13 2 13 3 11 2 03 3 04 3 11.5 1.54 01 1 0 11 1 1 12 1 1 11 2 0 11.5 1.7-1 -1

Sample Output

5-1

Source

Beijing 2004

提示

题意：

海底总动员的小鱼尼莫是一个熊孩子，独自进入深海区不慎进入了海底迷宫，不幸的是他忘记了回家的路，他给他爸爸马林发送求救信号，他爸看了看迷宫地图（不知哪来的），迷宫里有门和墙组成。马林不能穿墙，并且也要尽量少开门（可能因为开门的时候会遭到水母的毒）。请你计算出救出他儿子的最少开门次数。

思路：

这题从早上搞到晚上，思路是对的，但就是细节，要么没有初始化，要么。。。

这题比较难的是存图，存图，还是存图。。。把图存下来就可以开始BFS，我的方法是每个小方格作为坐标点，之后用结构体存下坐标以及该坐标上下左右的情况。还有注意尼莫在迷宫外的情况。

示例程序

Source CodeProblem: 2049Code Length: 3096BMemory: 1280KTime: 47MSLanguage: G++Result: Accepted#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct mp{    int up,down,l,r;}mp[201][201];struct nemo{    int x,y,step;}q[400000];int cmp(struct nemo a,struct nemo b){    if(a.step<b.step)    {        return 1;    }    else    {        return 0;    }}int bfs(int x,int y,int xmax,int ymax){    int v[201][201],f=0,top=0;    memset(v,0,sizeof(v));    q[top].x=y;//注意这里x和y进行了交换（周赛时忘了一直WA）    q[top].y=x;    q[top].step=0;    v[x][y]=1;    top++;    while(f<top)    {        if(q[f].x<1||q[f].y>ymax||q[f].x>xmax||q[f].y<1)        {            return q[f].step;        }        if(v[q[f].x-1][q[f].y]==0&&mp[q[f].x][q[f].y].down!=-1)        {            q[top].x=q[f].x-1;            q[top].y=q[f].y;            q[top].step=q[f].step+mp[q[f].x][q[f].y].down;//也注意一下这里加的是mp数组，有空地和门两种情况（周赛想都没想直接就加一了）            v[q[top].x][q[top].y]=1;            top++;        }        if(v[q[f].x+1][q[f].y]==0&&mp[q[f].x][q[f].y].up!=-1)        {            q[top].x=q[f].x+1;            q[top].y=q[f].y;            q[top].step=q[f].step+mp[q[f].x][q[f].y].up;            v[q[top].x][q[top].y]=1;            top++;        }        if(v[q[f].x][q[f].y+1]==0&&mp[q[f].x][q[f].y].r!=-1)        {            q[top].x=q[f].x;            q[top].y=q[f].y+1;            q[top].step=q[f].step+mp[q[f].x][q[f].y].r;            v[q[top].x][q[top].y]=1;            top++;        }        if(v[q[f].x][q[f].y-1]==0&&mp[q[f].x][q[f].y].l!=-1)        {            q[top].x=q[f].x;            q[top].y=q[f].y-1;            q[top].step=q[f].step+mp[q[f].x][q[f].y].l;            v[q[top].x][q[top].y]=1;            top++;        }        f++;        sort(q+f,q+top,cmp);//用优先队列来保证一定是开门次数最少    }    return -1;}int main(){    int n,m,x,y,d,t,i,i1,ymax,xmax;    double a,b;    scanf("%d %d",&n,&m);    while(n!=-1||m!=-1)    {        memset(mp,0,sizeof(mp));        ymax=-1;//对最大坐标进行初始化，后面需要判断尼莫是否在迷宫内        xmax=-1;        for(i=1;n>=i;i++)        {            scanf("%d %d %d %d",&x,&y,&d,&t);            if(d==0)            {                for(i1=0;t>i1;i1++)                {                    mp[y][x+i1].down=-1;//墙以-1来表示                    mp[y-1][x+i1].up=-1;                }                ymax=max(ymax,y);                xmax=max(xmax,x+i1);            }            else            {                for(i1=0;t>i1;i1++)                {                    mp[y+i1][x].l=-1;                    mp[y+i1][x-1].r=-1;                }                ymax=max(ymax,y+i1);                xmax=max(xmax,x);            }        }        for(i=1;m>=i;i++)        {            scanf("%d %d %d",&x,&y,&d);            if(d==0)            {                mp[y][x].down=1;//门以1来表示                mp[y-1][x].up=1;            }            else            {                mp[y][x].l=1;                mp[y][x-1].r=1;            }        }        scanf("%lf %lf",&a,&b);        x=a;        y=b;        if(x<1||y>ymax||x>xmax||y<1)        {            printf("0\n");//在外面就不要他爸去找了        }        else        {            printf("%d\n",bfs(x,y,xmax,ymax));        }        scanf("%d %d",&n,&m);    }    return 0;}