Codeforces 877D. Olya and Energy Drinks BFS+並查集

Olya loves energy drinks. She loves them so much that her room is full of empty cans from energy drinks.

Formally, her room can be represented as a field of n × m cells, each cell of which is empty or littered with cans.

Olya drank a lot of energy drink, so now she can run k meters per second. Each second she chooses one of the four directions (up, down, left or right) and runs from 1 to k meters in this direction. Of course, she can only run through empty cells.

Now Olya needs to get from cell (x1, y1) to cell (x2, y2). How many seconds will it take her if she moves optimally?

It’s guaranteed that cells (x1, y1) and (x2, y2) are empty. These cells can coincide.

Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 1000) — the sizes of the room and Olya’s speed.

Then n lines follow containing m characters each, the i-th of them contains on j-th position “#”, if the cell (i, j) is littered with cans, and “.” otherwise.

The last line contains four integers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — the coordinates of the first and the last cells.

Output
Print a single integer — the minimum time it will take Olya to get from (x1, y1) to (x2, y2).

If it’s impossible to get from (x1, y1) to (x2, y2), print -1. 題解：

感覺NOIP藥丸……這題都不會做……由於每個點只會被訪問一次，那麼我們只需要快速找到一個點上下左右四個方向沒有被訪問過的點就行了，這可以用並查集實現，剩下的就是常規的BFS了。 代碼：

#include<bits/stdc++.h>using namespace std;#define pa pair<int,int>#define LL long longconst int Maxn=1010;int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}    return x*f;}int k,n,m,stx,sty,edx,edy;int mp[Maxn][Maxn],num[Maxn][Maxn],nnum[Maxn*Maxn][2],tot=0;queue<int>qx,qy;bool vis[Maxn*Maxn];int f[Maxn*Maxn];int fa[4][Maxn*Maxn];//0123上下左右 int findfa(int o,int x){return((fa[o][x]==x)?x:fa[o][x]=findfa(o,fa[o][x]));}void del(int o,int x,int y){    fa[0][o]=findfa(0,fa[0][num[x-1][y]]);    fa[1][o]=findfa(1,fa[1][num[x+1][y]]);    fa[2][o]=findfa(2,fa[2][num[x][y-1]]);    fa[3][o]=findfa(3,fa[3][num[x][y+1]]);}int main(){    memset(f,63,sizeof(f));    n=read();m=read();k=read();    for(int i=1;i<=n;i++)    for(int j=1;j<=m;j++)    num[i][j]=++tot,nnum[tot][0]=i,nnum[tot][1]=j;    for(int i=1;i<=n;i++)    {        char str[Maxn];        scanf("%s",str+1);        for(int j=1;j<=m;j++)        {            mp[i][j]=((str[j]=='#')?0:1);            for(int l=0;l<4;l++)fa[l][num[i][j]]=num[i][j];        }    }    stx=read(),sty=read(),edx=read(),edy=read();    qx.push(stx);qy.push(sty);    int t=num[stx][sty];f[t]=0;vis[t]=true;    fa[0][t]=num[stx-1][sty];fa[1][t]=num[stx+1][sty];    fa[2][t]=num[stx][sty-1];fa[3][t]=num[stx][sty+1];    while(!qx.empty())    {        int x=qx.front(),y=qy.front();        if(x==edx&&y==edy)break;        qx.pop();qy.pop();        int t,o=num[x][y];        for(int i=0;i<4;i++)        {            while((t=findfa(i,fa[i][o]))!=0)            {                int tx=nnum[t][0],ty=nnum[t][1];                if(!mp[tx][ty])break;                if(i==0&&x-tx>k)break;                else if(i==1&&tx-x>k)break;                else if(i==2&&y-ty>k)break;                else if(i==3&&ty-y>k)break;                vis[t]=true;                f[t]=f[o]+1;                del(t,tx,ty);                qx.push(tx);qy.push(ty);            }        }    }    int ans=f[num[edx][edy]];    if(ans!=1061109567)printf("%d",ans);    else puts("-1");}