HDU1429胜利大逃亡(续)HDU 1885 Key Task BFS+状态压缩+水

HDU1429

只有10把钥匙 1<<10足够了

标记用三维数组

用Linux好不习惯 继续克服～

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAXN 11111
#include<queue>
const int INF = 999999;
char mp[33][33];
bool vis[33][33][1026];
int n,m;
struct node
{int x,y,step;int key;
};
queue<node>q;
int xx[4]= {1,0,-1,0};
int yy[4]= {0,1,0,-1};
int bfs(int x,int y)
{while(!q.empty())q.pop();memset(vis,false,sizeof(vis));node front,rear;front.x=x,front.y=y;front.key=0;front.step=0;vis[x][y][0]=true;q.push(front);while(!q.empty()){front=q.front();q.pop();for(int i=0; i<4; i++){int dx=front.x+xx[i],dy=front.y+yy[i];if(dx>=0&&dy>=0&&dx<n&&dy<m&&mp[dx][dy]!='*'&&!vis[dx][dy][front.key]){if(mp[dx][dy]=='^')//到终点{return front.step+1;}else if(mp[dx][dy]>='a'&&mp[dx][dy]<='z')//拿到钥匙{vis[dx][dy][front.key]=true;//标记rear.key =(front.key )| ( 1<<(mp[dx][dy]-'a'));//添加钥匙rear.x=dx,rear.y=dy,rear.step=front.step+1;q.push(rear);}else if(mp[dx][dy]>='A'&&mp[dx][dy]<='Z')//遇到门{if(front.key&(1<<(mp[dx][dy]-'A')))//如果有对应的钥匙{rear.x=dx,rear.y=dy,rear.key=front.key,rear.step=front.step+1;vis[dx][dy][rear.key]=true;q.push(rear);}}else{rear.x=dx,rear.y=dy,rear.key=front.key,rear.step=front.step+1;vis[dx][dy][rear.key]=true;q.push(rear);}}}}return 0;
}
int main()
{int t;//  freopen("in.txt","r",stdin);while(scanf("%d%d%d",&n,&m,&t)!=EOF){node s;for(int i=0; i<n; i++)scanf("%s",mp[i]);for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(mp[i][j]=='@')s.x=i,s.y=j,s.step=0;}}int ans=bfs(s.x,s.y);if(ans==0||ans>=t)printf("-1\n");else printf("%d\n",ans);}return 0;
}

HDU1885

这个就4把钥匙

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAXN 11111
#include<queue>
const int INF = 999999;
char mp[133][133];
bool vis[133][133][33];
int n,m;
int ke[28];
struct node
{int x,y,step;int key;
};
queue<node>q;
int xx[4]= {1,0,-1,0};
int yy[4]= {0,1,0,-1};
int bfs(int x,int y)
{while(!q.empty())q.pop();memset(vis,false,sizeof(vis));node front,rear;front.x=x,front.y=y;front.key=0;front.step=0;vis[x][y][0]=true;q.push(front);while(!q.empty()){front=q.front();q.pop();for(int i=0; i<4; i++){int dx=front.x+xx[i],dy=front.y+yy[i];if(dx>=0&&dy>=0&&dx<n&&dy<m&&mp[dx][dy]!='#'&&!vis[dx][dy][front.key]){if(mp[dx][dy]=='X')//到终点{return front.step+1;}else if(mp[dx][dy]>='a'&&mp[dx][dy]<='z')//拿钥匙{vis[dx][dy][front.key]=true;rear.key =(front.key )| ( 1<<(ke[mp[dx][dy]-'a']));rear.x=dx,rear.y=dy,rear.step=front.step+1;q.push(rear);}else if(mp[dx][dy]>='A'&&mp[dx][dy]<='Z')//开门{if(front.key&(1<<(ke[mp[dx][dy]-'A']))){rear.x=dx,rear.y=dy,rear.key=front.key,rear.step=front.step+1;vis[dx][dy][rear.key]=true;q.push(rear);}}else{rear.x=dx,rear.y=dy,rear.key=front.key,rear.step=front.step+1;vis[dx][dy][rear.key]=true;q.push(rear);}}}}return 0;
}
int main()
{//freopen("in.txt","r",stdin);ke['a'-'a']=0,ke['b'-'a']=1,ke['r'-'a']=2,ke['g'-'a']=3;while(scanf("%d%d",&n,&m),n+m){node s;for(int i=0; i<n; i++)scanf("%s",mp[i]);for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(mp[i][j]=='*')s.x=i,s.y=j,s.step=0;}}int ans=bfs(s.x,s.y);if(ans==0)printf("The poor student is trapped!\n");else printf("Escape possible in %d steps.\n",ans);}return 0;
}