zoj 1649 Rescue

 

又是类似骑士拯救公主，不过这个是朋友拯救天使的故事。。。

 

不同的是，天使有多个朋友，而骑士一般单枪匹马比较帅~

 

求到达天使的最短时间，杀死一个护卫1 units time ， 走一个格子 1 unit time 。SO，杀死一个护卫到达那个格子 2units time。

 

第一反应是广搜，就搜咧 = =。。

 

WA了，交hdu上 AC了，hdu数据真弱啊。。。

 

想了想，想通了，因为一般广搜的话必须都是1个时间才能搜，才能保证这个BFS树是等距离向外伸展的，而这个不是等距离的，所以需要一些处理。

 

1、我的方法是，找到天使后，把时间比下大小，最后输出最小的。需要优化，只这么做的话，会TLE的，如果走过一个格子，这个格子存走过时候的时间，下次再走到这个格子，如果时间比格子里的短，就入队，否则，就不用入队了。60MS。

 

2、网上看到另一种方法，就是把杀护卫和走到护卫那个格子看成两个动作等于说是入队两次，这个好啊！！！写了半天终于写出来了。20MS。

 

3、刚才想起来一种方法，因为如果等距离的BFS的话，队列里的time值是从小往大排的，那我直接用优先队列就可以了哈~~嘻嘻~10MS~人品好，爆0MS了~~

 

法I

 

#include <cstdio>#include <cstdlib>#include <iostream>#include <string.h>#include <queue>#include <limits.h>#define MAX 205using namespace std;typedef struct ANG{int t;int x,y;}ANG;queue <ANG> q;char map[MAX][MAX];int vis[MAX][MAX];int n,m,mmin;int dir[8] = {0,1,0,-1,1,0,-1,0};void BFS(){ANG tmp;int i,x,a,b,aa,bb,t;while( !q.empty() ){tmp = q.front();q.pop();a = tmp.x;b = tmp.y;t = tmp.t;for(i=0; i<8; i+=2){aa = a + dir[i];bb = b + dir[i+1];if( aa >= 0 && aa < n && bb >=0 && b < m && map[aa][bb] != '#' && vis[aa][bb] ){if( map[aa][bb] == 'a' ){if( t+1 < mmin )mmin = t+1;continue;}if( map[aa][bb] == '.' && t+1 < vis[aa][bb] ) //如果走过，而且时间比现在的短，就没必要入队了。 {vis[aa][bb] = t+1;tmp.t = t + 1;tmp.x = aa;tmp.y = bb;q.push(tmp);}if( map[aa][bb] == 'x' && t+2 < vis[aa][bb] ){vis[aa][bb] = t+2;tmp.t = t + 2;tmp.x = aa;tmp.y = bb;q.push(tmp);}}}}}int main(){int i,k,ans;ANG tmp;while( ~scanf("%d%d",&n,&m) ){memset(map,0,sizeof(map));for(i=0; i<n; i++)scanf("%s",map[i]);while( !q.empty() )q.pop();for(i=0; i<n; i++)for(k=0; k<m; k++)vis[i][k] = INT_MAX;for(i=0; i<n; i++)for(k=0; k<m; k++)if( map[i][k] == 'r' ){map[i][k] = '.';tmp.x = i;tmp.y = k;tmp.t = 0;q.push(tmp);vis[i][k] = 0;}mmin = INT_MAX;BFS();if( mmin != INT_MAX )printf("%d/n",mmin);elseprintf("Poor ANGEL has to stay in the prison all his life./n");}return 0;} 

 

法II

 

#include <cstdio>#include <cstdlib>#include <iostream>#include <string.h>#include <queue>#include <limits.h>#define MAX 201using namespace std;typedef struct ANG{int t;int x,y,flag;}ANG;char map[MAX][MAX];int vis[MAX][MAX];int n,m;int dir[8] = {0,1,0,-1,1,0,-1,0};priority_queue<ANG> q;bool operator<(ANG a, ANG b){return a.t > b.t;}int BFS(){ANG tmp;int i,x,a,b,aa,bb,t;while( !q.empty() ){tmp = q.top();q.pop();a = tmp.x;b = tmp.y;t = tmp.t;if( map[a][b] == 'x' && tmp.flag == 1 ){tmp.x = a;tmp.y = b;tmp.t = t+1;tmp.flag = 2;q.push(tmp);vis[a][b] = 1;continue;}for(i=0; i<8; i+=2){aa = a + dir[i];bb = b + dir[i+1];if( aa >= 0 && aa < n && bb >=0 && b < m && map[aa][bb] != '#' && !vis[aa][bb] ){if( map[aa][bb] == 'a' )return t+1;if( map[aa][bb] == '.' ) {vis[aa][bb] = 1;tmp.t = t + 1;tmp.x = aa;tmp.y = bb;q.push(tmp);}if( map[aa][bb] == 'x' ){vis[aa][bb] = 0;tmp.flag = 1;tmp.t = t + 1;tmp.x = aa;tmp.y = bb;q.push(tmp);}}}}return -1;}int main(){int i,k,ans;ANG tmp;while( ~scanf("%d%d",&n,&m) ){memset(map,0,sizeof(map));for(i=0; i<n; i++)scanf("%s",map[i]);while( !q.empty() )q.pop();memset(vis,0,sizeof(vis));for(i=0; i<n; i++)for(k=0; k<m; k++)if( map[i][k] == 'r' ){map[i][k] = '.';tmp.x = i;tmp.y = k;tmp.t = 0;q.push(tmp);vis[i][k] = 2;}ans = BFS();if( ans != -1 )printf("%d/n",ans);elseprintf("Poor ANGEL has to stay in the prison all his life./n");}return 0;} 

 

法III

 

#include <cstdio>#include <cstdlib>#include <iostream>#include <string.h>#include <queue>#include <limits.h>#define MAX 201using namespace std;typedef struct ANG{int t;int x,y;}ANG;char map[MAX][MAX];int vis[MAX][MAX];int n,m;int dir[8] = {0,1,0,-1,1,0,-1,0};priority_queue<ANG> q;bool operator<(ANG a, ANG b){return a.t > b.t;}int BFS(){ANG tmp;int i,x,a,b,aa,bb,t;while( !q.empty() ){tmp = q.top();q.pop();a = tmp.x;b = tmp.y;t = tmp.t;for(i=0; i<8; i+=2){aa = a + dir[i];bb = b + dir[i+1];if( aa >= 0 && aa < n && bb >=0 && b < m && map[aa][bb] != '#' && !vis[aa][bb] ){vis[aa][bb] = 1;if( map[aa][bb] == 'a' )return t+1;if( map[aa][bb] == '.' ) {tmp.t = t + 1;tmp.x = aa;tmp.y = bb;q.push(tmp);}if( map[aa][bb] == 'x' ){tmp.t = t + 2;tmp.x = aa;tmp.y = bb;q.push(tmp);}}}}return -1;}int main(){int i,k,ans;ANG tmp;while( ~scanf("%d%d",&n,&m) ){memset(map,0,sizeof(map));for(i=0; i<n; i++)scanf("%s",map[i]);while( !q.empty() )q.pop();memset(vis,0,sizeof(vis));for(i=0; i<n; i++)for(k=0; k<m; k++)if( map[i][k] == 'r' ){map[i][k] = '.';tmp.x = i;tmp.y = k;tmp.t = 0;q.push(tmp);vis[i][k] = 1;}ans = BFS();if( ans != -1 )printf("%d/n",ans);elseprintf("Poor ANGEL has to stay in the prison all his life./n");}return 0;}