http://poj.org/problem?id=3026
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1248
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Borg Maze
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 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 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, 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 Source |
题意：
给出一个迷宫，‘#’是墙壁，‘ ’（空格）可走，‘S’是起点，‘A’是目标，一个群体从S点开始，每次可以走周围相邻的4个格子，走到某个目标的花费是从上一个目标（或起点）开始计算的步数，群体可且仅可在S或A出分成若干个（可以看成是无数个，即使在同一个格子中）群体。比如从S开始走5步到A1,在A1分成两个群体，其中一个到达A2走3步，另一个到达A3也走3步，那么总花费是5+3+3=11。求到达所有A的最小花费。
分析：
题意很难理解，其实就是个最小生成树，用BFS在平面内模拟prim算法即可，这里要用到优先队列，因为到达某个A后就相当于步数清0,每次选走过步数最少的扩展状态，注意格子是可以重复走的，但只有步数比之前到达这个格子的步数少才走。
POJ数据好恶心，每行后面还有一堆空格，UVA就没这种情况。
/*
*
* Author : fcbruce
*
* Date : 2014-08-11 15:26:39
*
*/
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm
#define maxn
using namespace std;
struct node
{
int x,y,w;
bool operator < (const node &n)const
{
return w>n.w;
}
};
int n,m,total;
char G[101][101];
char buf[101];
int vis[101][101];
int sx,sy;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
int bfs()
{
int res=0;
priority_queue<node > q;
q.push( (node){sx,sy,0});
memset(vis,0x3f,sizeof vis);
vis[sx][sy]=0;
while (!q.empty() || total)
{
node s=q.top();q.pop();
if (G[s.x][s.y]=='A')
{
res+=s.w;
s.w=0;
G[s.x][s.y]=' ';
total--;
}
for (int i=0;i<4;i++)
{
int tx=s.x+dx[i];
int ty=s.y+dy[i];
int tw=s.w+1;
if (G[tx][ty]!='#' && vis[tx][ty]>tw)
{
q.push( (node){tx,ty,tw});
vis[tx][ty]=tw;
}
}
}
return res;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("/home/fcbruce/文档/code/t","r",stdin);
#endif // ONLINE_JUDGE
int T_T;
scanf( "%d",&T_T);
while (T_T--)
{
scanf( "%d %d",&m,&n);gets(buf);
for (int i=0;i<n;i++)
gets(G[i]);
total=0;
for (int i=0;i<n;i++)
{
for (int j=0;j<m;j++)
{
if (G[i][j]=='S')
{
sx=i;sy=j;
}
if (G[i][j]=='A')
total++;
}
}
printf( "%d\n",bfs());
}
return 0;
}
POJ 3026 Borg Maze & UVA 10307 Killing Aliens in Borg Maze(BFS,最小生成树)