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  • bzoj:1656 [Usaco2006 Jan] The Grove 树木

    Description

    The pasture contains a small, contiguous grove of trees that has no 'holes' in the middle of the it. Bessie wonders: how far is it to walk around that grove and get back to my starting position? She's just sure there is a way to do it by going from her start location to successive locations by walking horizontally, vertically, or diagonally and counting each move as a single step. Just looking at it, she doesn't think you could pass 'through' the grove on a tricky diagonal. Your job is to calculate the minimum number of steps she must take. Happily, Bessie lives on a simple world where the pasture is represented by a grid with R rows and C columns (1 <= R <= 50, 1 <= C <= 50). Here's a typical example where '.' is pasture (which Bessie may traverse), 'X' is the grove of trees, '*' represents Bessie's start and end position, and '+' marks one shortest path she can walk to circumnavigate the grove (i.e., the answer): ...+... ..+X+.. .+XXX+. ..+XXX+ ..+X..+ ...+++* The path shown is not the only possible shortest path; Bessie might have taken a diagonal step from her start position and achieved a similar length solution. Bessie is happy that she's starting 'outside' the grove instead of in a sort of 'harbor' that could complicate finding the best path.

    牧场里有一片树林,林子里没有坑.
        贝茜很想知道,最少需要多少步能围绕树林走一圈,最后回到起点.她能上下左右走,也能走对角线格子.牧场被分成R行C列(1≤R≤50,1≤C≤50).下面是一张样例的地图,其中“.”表示贝茜可以走的空地,  “X”表示树林,  “*”表示起点.而贝茜走的最近的路已经特别地用“+”表示出来.
     
     
     
     
     
    题目保证,最短的路径一定可以找到.

    Input

    * Line 1: Two space-separated integers: R and C

    * Lines 2..R+1: Line i+1 describes row i with C characters (with no spaces between them).

        第1行输入R和C,接下来R行C列表示一张地图.地图中的符号如题干所述.

    Output

    * Line 1: The single line contains a single integer which is the smallest number of steps required to circumnavigate the grove.

        输出最少的步数.

    Sample Input

    6 7
    .......
    ...X...
    ..XXX..
    ...XXX.
    ...X...
    ......*

    Sample Output

    13

    随便找棵树然后画一条射线作为分界线,使之不能通过,然后就行BFS啦

    以样例为例:

    在第二行唯一一棵树那划条射线,然后BFS结果如下:(-1为树木,0为起点)

    在这个数据里面并不明显,划过线的部分是不能通过的……

    再看这个数据:

     

    结果如下:

    其他看代码咯

    #include<queue>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    
    struct na{
        int x,y;
    };
    int n,m,x=-1,y;
    const int fx[8]={0,0,1,-1,1,1,-1,-1},fy[8]={1,-1,0,0,1,-1,1,-1};
    int map[51][51];
    char c;
    queue <na> q;
    int main(){
        scanf("%d%d",&n,&m);
        for (int i=0;i<n;i++)
        for (int j=0;j<m;j++){
            c=getchar();
            while(c=='
    ') c=getchar();
            if (c=='X') {
                map[i][j]=-1;
                if (x==-1) x=i,y=j;
            }else
            if (c=='*'){
                na tmp;tmp.x=i;tmp.y=j;q.push(tmp);
            }else map[i][j]=2500;
        }
        while(!q.empty()){
            na k=q.front();q.pop();
            for (int i=0;i<8;i++){
                na now=k;
                now.x+=fx[i];now.y+=fy[i];
                if (now.x<0||now.y<0||now.x>=n||now.y>=m) continue;
                if (k.y<=y&&k.x==x&&now.x==x-1) continue;
                if (k.y<=y&&k.x==x-1&&now.x==x) continue;
                if (map[now.x][now.y]>map[k.x][k.y]+1){
                    map[now.x][now.y]=map[k.x][k.y]+1;
                    q.push(now);
                }
            }
        }
        int ans=2500;
        for (int i=y-1;i>=0;i--){
            if (map[x][i]+map[x-1][i]<ans) ans=map[x][i]+map[x-1][i];
            if (map[x][i]+map[x-1][i+1]<ans) ans=map[x][i]+map[x-1][i+1];
            if (i) if (map[x][i]+map[x-1][i-1]<ans) ans=map[x][i]+map[x-1][i-1];
        }
        printf("%d
    ",ans+1);
    }
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  • 原文地址:https://www.cnblogs.com/Enceladus/p/5020280.html
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