URAL1325-Dirt

Dirt

— Hello, may I speak to Petrov, please? Hello, my darling… You know, there was a little accident at our home… No, no, don’t worry, your computer was not damaged. It is only a bit dirty there now. Well, I should say it’s very dirty there and I’m at my Mom’s now. Of course, I’ll clean it… When? Well, maybe when I have my vacation. What? Well, when we are back from Turkey… the next vacation then. I’ll stay at Mother’s until then, and you may live here also. No, no, I don’t insist, sure, you may stay at home if you wish so. I prepared boots for you, they are at the door. But please, don’t make it worse, before you step on a clean floor, change your boots, put on your slippers, they are at the door also. Take them with you when you walk through the dirt. And when you walk on a clean floor, take the boots with you. You see, the dirt is in different places. OK, my love? Thank you!
It is not a great pleasure to change boots each time you get from a clean floor to a dirty floor and vice versa, it’s easier to walk extra several meters. So it is necessary to find a way of getting from one place in the apartment to another with the minimal possible number of boots changes; and among these paths the shortest one must be found.
To begin with, it is natural to determine an optimal way of passing the Most Important Route: from the computer to the refrigerator.
Input
The first line of the input contains two integers M and N, which are dimensions of the apartment (in meters), 1 <= N, M <= 500. The two integers in the second line are the coordinates of the computer, and the third line contains the coordinates of the refrigerator. Each of the following M lines contains N symbols; this is the plan of the apartment. On the plan, 1 denotes a clean square, 2 denotes a dirty square, and 0 is either a wall or a square of impassable dirt. It is possible to get from one square to another if they have a common vertex. When you pass from a clean square to a dirty one or vice versa, you must change shoes. The computer and the refrigerator are not on the squares marked with 0.
The upper left square of the plan has coordinates (1, 1).
Output
You should output two integers in one line separated with a space. The first integer is the length of the shortest path (the number of squares on this path including the first and the last squares) with the minimal possible number of boots changes. The second number is the number of boots changes. If it is impossible to get from the computer to the refrigerator, you should output 0 0.
Example
input
3 7
1 1
3 7
1200121
1212020
1112021
output
8 4

题目大意： 有一张N*M的地图，上面有些格子是干净的有些是脏的，人从干净的格子走到脏的格子需要换鞋，从脏的走到干净的也需要换鞋。一个人从给定起点走到给定终点，问在换鞋次数最小的情况下最短距离是多少？
解题思路： 建图，将地图中每个格子作为点，相邻的非障碍物格子之间连边，用优先队列BFS，求出到达每个格子的最小换鞋数和步数，最后看终点的步数是否是INF，若是，则不可达，否则输出相应的结果。时间复杂度O(NM)O(NM)
注意点：这里的相邻是八个方向！

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;
typedef long long LL;
const int MAXN=505;
const int INF=0x3f3f3f3f;
int sx,sy,ex,ey;
char G[MAXN][MAXN];
int dx[8]={0,0,1,-1,1,1,-1,-1};
int dy[8]={1,-1,0,0,1,-1,1,-1};
bool vis[MAXN][MAXN];
int n,m;
int ans1,ans2;struct node
{int x,y;int step;int change;/*node(int _x,int _y,int _step,int _change){x=_x;y=_y;step=_step;change=_change;}*/friend bool operator < (node a,node b){if(a.change==b.change) return a.step>b.step;return a.change>b.change;}
}grid[MAXN][MAXN];void bfs(int sx,int sy)
{vis[sx][sy]=true;grid[sx][sy].step=1;grid[sx][sy].change=0;priority_queue<node> pq;pq.push(grid[sx][sy]);while(!pq.empty()){node fst=pq.top();pq.pop();/*if(fst.x==ex-1&&fst.y==ey-1){ans1=fst.step;ans2=fst.change;break;}*/for(int i=0;i<8;i++){int tx=fst.x+dx[i];int ty=fst.y+dy[i];if(tx<0||tx>=n||ty<0||ty>=m||G[tx][ty]=='0') continue;vis[tx][ty]=true;node nxt;nxt.x=tx;nxt.y=ty;if(G[tx][ty]!=G[fst.x][fst.y])nxt.change=fst.change+1;elsenxt.change=fst.change;nxt.step=fst.step+1;if(grid[tx][ty]<nxt){grid[tx][ty]=nxt;pq.push(nxt);}}}
}int main()
{ios::sync_with_stdio(false);while(cin>>n>>m){cin>>sx>>sy>>ex>>ey;for(int i=0;i<n;i++){for(int j=0;j<m;j++){vis[i][j]=false;cin>>G[i][j];grid[i][j].x=i;grid[i][j].y=j;grid[i][j].change=INF;grid[i][j].step=INF;}}bfs(sx-1,sy-1);if(vis[ex-1][ey-1])cout<<grid[ex-1][ey-1].step<<" "<<grid[ex-1][ey-1].change<<endl;elsecout<<0<<" "<<0<<endl;}return 0;
}

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