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快速切题 poj3026

2024-07-17 08:12:46   219人阅读

感受到出题人深深的~恶意

这提醒人们以后。。。数字后面要用gets~不要getchar

此外。。不要相信那个100？

Borg Maze
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8524 Accepted: 2872

Description

The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated subspace network that insures each member is given constant supervision and guidance. 

Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.

Input

On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character, a space `` ‘‘ stands for an open space, a hash mark ``#‘‘ stands for an obstructing wall, the capital letter ``A‘‘ stand for an alien, and the capital letter ``S‘‘ stands for the start of the search. The perimeter of the maze is always closed, i.e., there is no way to get out from the coordinate of the ``S‘‘. At most 100 aliens are present in the maze, and everyone is reachable.

Output

For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.

Sample Input

26 5##### #A#A### # A##S  ####### 7 7#####  #AAA####    A## S ####     ##AAA########

Sample Output

811
思路： bfs遍历得两点距（这道题数据不大），prim即可
应用时：30min
实际用时：2h
//M: 5056K T:47MS#include<cstdio>#include <cstring>#include <queue>#include <assert.h>using namespace std;char maz[202][202];//迷宫bool vis[202][202];//用来存储bfs过程中是否遍历以节约时间int d[201][202][202];//d[aliennum][y][x]从第alinum到坐标y,x的最小步数 int alien[201][2];//记录alien坐标,地球人也是alien之一int numa;//alien数目int x,y;//迷宫总长int sx,sy;//人类起点int e[202][202];//建图typedef pair<int,int> P;queue<P>  que;//用来存储bfs结果const int dx[4]={0,0,-1,1};const int dy[4]={-1,1,0,0};bool used[101];//用来存储prim过程void bfs(){    sx=sy=-1;    numa=0;    memset(vis,0,sizeof(vis));    for(int i=0;i<y;i++){//为了把s点作为起点所以特意寻找,其实不用        for(int j=0;j<x;j++){            if(maz[i][j]==‘S‘){                sx=j;                sy=i;                break;            }        }    }    assert(sx!=-1&&sy!=-1);//判断有没有找到sxsy,这次用这个发现scanf读迷宫不可    vis[sy][sx]=true;//起始点当然已经走过了    d[0][sy][sx]=0;    alien[numa][0]=sy;    alien[numa++][1]=sx;    que.push(P(sy,sx));//分层bfs    while(!que.empty()){        P p=que.front();        que.pop();        for(int i=0;i<4;i++){            int ny=p.first+dy[i];            int nx=p.second+dx[i];            if(nx>=0&&nx<x&&ny>=0&&ny<y&&!vis[ny][nx]&&maz[ny][nx]!=‘#‘){                d[0][ny][nx]=d[0][p.first][p.second]+1;                vis[ny][nx]=true;                que.push(P(ny,nx));                if(maz[ny][nx]==‘A‘){//把所有找到的外星人编个号                    alien[numa][0]=ny;                    alien[numa++][1]=nx;                }            }        }    }    for(int ai=1;ai<numa;ai++){//以各个外星人作为起点开始        memset(vis,0,sizeof(vis));        int ay=alien[ai][0],ax=alien[ai][1];//外星人地址        d[ai][ay][ax]=0;        que.push(P(ay,ax));        vis[ay][ax]=true;//起点当然已经遍历        while(!que.empty()){            P p=que.front();            que.pop();            for(int i=0;i<4;i++){                int ny=p.first+dy[i];                int nx=p.second+dx[i];                if(nx>=0&&nx<x&&ny>=0&&ny<y&&!vis[ny][nx]&&maz[ny][nx]!=‘#‘){                    d[ai][ny][nx]=d[ai][p.first][p.second]+1;                    vis[ny][nx]=true;                    que.push(P(ny,nx));//从这一点出发的状况加入队列                }            }        }    }    for(int i=0;i<numa;i++){//取边建图        for(int j=0;j<numa;j++){            int ay1=alien[j][0],ax1=alien[j][1];//所有的其他外星人坐标,自己对自己肯定是0            e[i][j]=d[i][ay1][ax1];        }    }}priority_queue <P,vector<P>, greater <P> > pque;int prim(){    memset(used,0,sizeof(used));//多组case所以memset    used[0]=true;//alien[0]是人类,不过也无所谓,总之这个点走过了    int unum=1;//这时候已经加入了一个点,就是人类起点,用来存储当前树有多少已加入节点    for(int i=1;i<numa;i++){        pque.push(P(e[0][i],i));//把到当前树的所有边都加入以找到最小边,这条边所连的那个点进树没有更短的路径了,因为是树所以这步贪心只会优化结果不会丢失结果    }    int ans=0;    while(unum<numa){        int t=pque.top().second;        int td=pque.top().first;        pque.pop();        if(used[t])continue;        ans+=td;        used[t]=true;        unum++;        for(int i=0;i<numa;i++){            if(!used[i]){                pque.push(P(e[t][i],i));//按边排序            }        }    }    while(!pque.empty())pque.pop();//多组case所以pop,也许直接建立在过程里更好吧    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d%d",&x,&y);        gets(maz[0]);//不想再建一个专门接收空格的字符串所以用了maz[0]        for(int i=0;i<y;i++){            gets(maz[i]);        }        bfs();//取点建图        int ans=prim();        printf("%d\n",ans);    }}

　　


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