Problem Description
Angel was caught by the MOLIGPY! He was put in * by Moligpy. The * is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the *.
Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?
Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?
) to move into the grid. We assume that we moving up,
down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.
You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
Input
First line contains two integers stand for N and M.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
Process to the end of the file.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
Process to the end of the file.
Output
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the * all his life."
Sample Input
7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........
Sample Output
13 题意：一个牢房图。#是墙壁，a是angel。r是angel的朋友。x是敌人，每走一步消耗1个单位时间。消灭1个敌人也消耗一个单位时间。求r到a的最小时间。 。 依照正常的思路：直接从R BFS 到 a 遇到 x就多加一个时间 即可了 只是这样是不正确了（题目太水 还是A了）
伪AC代码：#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
#define w 205
char map[w][w];
int vis[w][w];
int sx,sy;
int m,n;
struct node {
int x,y,time;
};
int fx[4][2]={1,0,0,1,-1,0,0,-1};
int bfs()
{
int tx,ty;
memset(vis,0,sizeof(vis));
node now,next;
queue<node>q;
now.x=sx;now.y=sy;now.time=0;
vis[sx][sy]=1;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
if(map[now.x][now.y]=='r')
return now.time;
for(int i=0;i<4;i++)
{
tx=now.x+fx[i][0];
ty=now.y+fx[i][1];
if(tx<1||tx>m||ty<1||ty>n||vis[tx][ty]==1||map[tx][ty]=='#')
continue;
vis[tx][ty]=1;
next.x=tx;
next.y=ty;
if(map[tx][ty]=='x')
next.time=now.time+2;
else
next.time=now.time+1;
q.push(next);
}
}
return -1;
}
int main()
{
int i,j;
while(cin>>m>>n)
{
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
{
cin>>map[i][j];
if(map[i][j]=='a')
{
sx=i;sy=j;
}
}
int ss= bfs();
if(ss==-1)
printf("Poor ANGEL has to stay in the * all his life.\n");
else
printf("%d\n",ss);
}
return 0;
}但面对这组数据：
3 4a...
##x.
###r结果 却是6（应该是5啊）
原因事实上是（2,3） 和（1.4）是同步进入队列的 进入的顺序和方向数组有关= =
这个问题能够用优先队列解决。 详见代码
优先队列版：#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
#define w 205
char map[w][w];
int vis[w][w];
int sx,sy;
int m,n;
struct node {
int x,y,time;
bool operator <(const node & t) const
{
return time>t.time; //改成<号 则较大的先出队
}
};
int fx[4][2]={1,0,0,1,-1,0,0,-1};
int bfs()
{
int tx,ty;
memset(vis,0,sizeof(vis));
node now,next;
priority_queue<node>q; //加上前缀 priority_
now.x=sx;now.y=sy;now.time=0;
vis[sx][sy]=1;
q.push(now);
while(!q.empty())
{
now=q.top(); //优先队列不能用 q.front();
q.pop();
if(map[now.x][now.y]=='r')
return now.time;
for(int i=0;i<4;i++)
{
tx=now.x+fx[i][0];
ty=now.y+fx[i][1];
if(tx<1||tx>m||ty<1||ty>n||vis[tx][ty]==1||map[tx][ty]=='#')
continue;
vis[tx][ty]=1;
next.x=tx;
next.y=ty;
if(map[tx][ty]=='x')
next.time=now.time+2;
else
next.time=now.time+1;
q.push(next);
}
}
return -1;
}
int main()
{
int i,j;
while(cin>>m>>n)
{
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
{
cin>>map[i][j];
if(map[i][j]=='a')
{
sx=i;sy=j;
}
}
int ss= bfs();
if(ss==-1)
printf("Poor ANGEL has to stay in the * all his life.\n");
else
printf("%d\n",ss);
}
return 0;
}模板。
。。
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