Borg Maze

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8905   Accepted: 2969

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

2
6 5
#####
#A#A##
# # A#
#S ##
#####
7 7
#####
#AAA###
# A#
# S ###
# #
#AAA###
#####

Sample Output

8
11 思路:bfs得外星人间最小距离(假设人类是第0号外星人),prim得最小生成树
#include<cstdio>
#include <cstring>
#include <queue>
#include <assert.h>
using namespace std;
char maz[202][202];
bool vis[202][202];
int d[201][202][202];
int alien[201][2];
int numa;
int x,y;
int sx,sy;
int e[202][202];
typedef pair<int,int> P;
queue<P> que;
const int dx[4]={0,0,-1,1};
const int dy[4]={-1,1,0,0};
bool used[101];
void bfs(){
sx=sy=-1;//处理迷宫
numa=0;
memset(vis,0,sizeof(vis));
for(int i=0;i<y;i++){
for(int j=0;j<x;j++){
if(maz[i][j]=='S'){
sx=j;
sy=i;
break;
}
}
}
assert(sx!=-1&&sy!=-1);
vis[sy][sx]=true;
d[0][sy][sx]=0;
alien[numa][0]=sy;
alien[numa++][1]=sx;
que.push(P(sy,sx));
while(!que.empty()){//从人类出发bfs
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'){
alien[numa][0]=ny;
alien[numa++][1]=nx;
}
}
}
}
for(int ai=1;ai<numa;ai++){//从每个外星人出发bfs
memset(vis,0,sizeof(vis));
int ay=alien[ai][0],ax=alien[ai][1];
d[ai][ay][ax]=0;
que.push(P(ay,ax));
vis[ay][ax]=true;
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));
}
}
}
}
for(int i=0;i<numa;i++){//建图
//int ay=alien[i][0],ax=alien[i][1];
for(int j=0;j<numa;j++){
int ay1=alien[j][0],ax1=alien[j][1];
e[i][j]=d[i][ay1][ax1];
}
}
}
priority_queue <P,vector<P>, greater <P> > pque;
int prim(){
memset(used,0,sizeof(used));
used[0]=true;
int unum=1;
for(int i=1;i<numa;i++){
pque.push(P(e[0][i],i));
}
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));
}
}
}
while(!pque.empty())pque.pop();
return ans;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&x,&y);
gets(maz[0]);//抛弃空行
for(int i=0;i<y;i++){
gets(maz[i]);
}
bfs();
int ans=prim();
printf("%d\n",ans);
}
}

  

快速切题 poj 3026 Borg Maze 最小生成树+bfs prim算法 难度:0的更多相关文章

  1. poj 3026 Borg Maze (最小生成树+bfs)

    有几个错误,调试了几个小时,样例过后 1Y. 题目:http://poj.org/problem?id=3026 题意:就是让求A们和S的最小生成树 先用bfs找每两点的距离,再建树.没剪枝 63MS ...

  2. poj 3026 Borg Maze 最小生成树 + 广搜

    点击打开链接 Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7097   Accepted: 2389 ...

  3. POJ 3026 Borg Maze【BFS+最小生成树】

    链接: http://poj.org/problem?id=3026 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#probl ...

  4. POJ 3026 Borg Maze(bfs+最小生成树)

    Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6634   Accepted: 2240 Descrip ...

  5. POJ 3026 --Borg Maze(bfs,最小生成树,英语题意题,卡格式)

    Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16625   Accepted: 5383 Descri ...

  6. 快速切题 poj 2993 Emag eht htiw Em Pleh 模拟 难度:0

    Emag eht htiw Em Pleh Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2806   Accepted:  ...

  7. poj 3026 Borg Maze (BFS + Prim)

    http://poj.org/problem?id=3026 Borg Maze Time Limit:1000MS     Memory Limit:65536KB     64bit IO For ...

  8. POJ 3026 : Borg Maze(BFS + Prim)

    http://poj.org/problem?id=3026 Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions ...

  9. POJ - 3026 Borg Maze BFS加最小生成树

    Borg Maze 题意: 题目我一开始一直读不懂.有一个会分身的人,要在一个地图中踩到所有的A,这个人可以在出发地或者A点任意分身,问最少要走几步,这个人可以踩遍地图中所有的A点. 思路: 感觉就算 ...

随机推荐

  1. spoj8222

    地址: 题目: NSUBSTR - Substrings no tags  You are given a string S which consists of 250000 lowercase la ...

  2. hive union all使用注意

    UNION用于联合多个select语句的结果集,合并为一个独立的结果集,结果集去重. UNION ALL也是用于联合多个select语句的结果集.但是不能消除重复行.现在hive只支持UNION AL ...

  3. 20135302魏静静——linux课程第八周实验及总结

    linux课程第八周实验及总结 实验及学习总结 1. 进程切换在内核中的实现 linux中进程切换是很常见的一个操作,而这个操作是在内核中实现的. 实现的时机有以下三个时机: 中断处理过程(包括时钟中 ...

  4. 重新想,重新看——CSS3变形,过渡与动画①

    学习CSS3,觉得最难记忆的部分除了flex特性之外,就要属变形,过渡和动画部分了.作为初学者,总有种犯懒的心理,想着既然IE8浏览器都不完全支持CSS动画属性,还要考虑浏览器兼容问题,那么就不那么着 ...

  5. gerrit代码审核工具之“error unpack failed error Missing unknown”错误解决思路

    使用gerrit代码审核工具时遇到error: unpack failed: error Missing unknown d6d7c89bd1d77f44c5c8e99437aaffbfc0684e7 ...

  6. centos配置yum源为中国镜像源

    有时候CentOS默认的yum源不一定是国内镜像,导致yum在线安装及更新速度不是很理想.这时候需要将yum源设置为国内镜像站点.国内主要开源的开源镜像站点应该是网易和阿里云了. 修改CentOS默认 ...

  7. 剑指Offer——最长不包含重复字符的子字符串

    Solution 动态规划. f(i)表示包含第i个字符的最长子串. 如果第i个字符没在之前出现过,那么f(i) = f(i - 1) + 1 如果第i个字符在之前出现过,这个时候应该分两种情况,假设 ...

  8. 【异常记录(五)】C# 无法发送具有此谓词类型的内容正文错误

    今天请求接口直接调了以前写好的方法,结果报了(405)不支持方法的错误,一看是GET写成POST了,改成GET之后,又报了无法发送具有此谓词类型的内容正文错误的错误 原来之前的方法里面有GetRequ ...

  9. C#之多线程

    多线程在C#中使用得非常频繁,线程之间的充分利用显得尤为重要,一般的写法都是得不到充分利用资源,本人针对多线程写了一种方法,可以充分利用资源,保证每次同时启动10条线程,现在执行完马上再启动一条,总之 ...

  10. 在lnmp下开启fileinfo扩展 Ubuntu系统

    在lnmp下开启fileinfo扩展 Ubuntu系统 1.进入目录下 cd /usr/local/lnmp1.4-full/src/php-5.6.31/ext/fileinfo 2.phpize处 ...