HDU 1241 HDU 1495 HDU2612

蒻苣：弱弱很弱，路过的巨巨还不吝赐教！^.^...QAQ

三题都分类为搜索。。水搜索。。。

不过HDU 1495 貌似模拟也行的。。。

HDU 1241 Oil Deposits

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1241

题意：给出一个n*m的地图，求油田的数量。 其中 上下等八个方向只要相连就算作一个油田。

题解：BFS或者DFS。 数据不大，都可以做。

注意：题面输入的地方有个坑，就是输入n和m后面有个空格。。用scanf的小心。。。。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 char s[200+5][200+5];
 8 int vis[200+5][200+5];
 9 int dir[8][2] = {1,1,-1,-1,1,-1,-1,1,1,0,-1,0,0,1,0,-1};
10 struct st{
11         int x,y;
12 };
13 queue<st>q;
14 void bfs(int x,int y,int n,int m)
15 {
16        st f;
17        f.x = x;
18        f.y = y;
19        vis[x][y] = 1;
20        q.push(f);
21        while( !q.empty() )
22        {
23                f = q.front();
24                st g;
25                q.pop();
26                for(int i = 0;i < 8; i++)
27                {
28                        g.x = f.x + dir[i][0];
29                        g.y = f.y + dir[i][1];
30                        if(!vis[g.x][g.y]&&s[g.x][g.y]=='@')
31                        {
32                                q.push(g);
33                                vis[g.x][g.y] = 1;
34                        }
35                }
36        }
37 }
38 int main()
39 {
40         int n,m;
41         while(cin>>n>>m)
42         {
43                 if(!m) break;
44                 memset(vis,0,sizeof(vis));
45                 memset(s,0,sizeof(s));
46                 int x1,x2,y1,y2;
47                 for(int i = 1; i <= n; i++ ) cin>>s[i]+1;
48                 int sum  = 0;
49                 for(int i = 1;i <= n; i++)
50                         for(int j = 1; j <= m;j++ )
51                                 if(s[i][j] == '@' && !vis[i][j])
52                                 {bfs(i,j,n,m);sum++;}
53                 printf("%d\n",sum);
54         }
55         return 0;
56 }
View Code

HDU 1495 非常可乐

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1495

题意：中文题。。。

题解：模拟或者搜索都行吧。。模拟 0ms过的 搜索 15ms（其实也差不多是0）

模拟代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int sum;
 7 void dfs(int m,int n,int k)
 8 {
 9     if(n<k)
10         swap(n,k);
11     int a,b,c;
12     sum = 1;
13     b = n;
14     c = 0;
15     a = k;
16     while(a!=m)
17     {
18         if(c==0)
19         {
20             if(b>=k)
21             {
22                 c=k;
23                 b-=c;
24             }
25             else
26             {
27                 c=b;
28                 b=0;
29             }
30             sum++;
31         }
32         else
33         {
34             if(c==k)
35             {
36                 a+=c;
37                 c=0;
38                   sum++;                                
39             }
40             else if( b==0)
41             {
42                 b=n;
43                 a-=b;
44                 sum++;
45             }
46             else if(b>0)
47             {
48                 if(b>=k-c)
49                 {
50                     int t = k-c;
51                     c=k;
52                     b-=t;
53                 }
54                 else
55                 {
56                     b=0;
57                     c+=b;
58                 }
59                 sum++;
60             }
61         }
62         if(a==b+c)
63             break;
64     }
65     if(a == m)
66         sum = 0;
67 }
68 
69 int main()
70 {
71     int m,n,k;
72     while(cin>>m>>n>>k)
73     {
74         sum = 0;
75         if( m == 0 && n == 0 && k == 0)
76          break;
77         if(n+k<m)
78          cout<<"NO"<<endl;
79         else if(n==k)
80          cout<<"1"<<endl;
81         else{
82             dfs(m,n,k);
83             if(sum)
84                 cout<<sum<<endl;
85             else
86                 cout<<"NO"<<endl;
87         }
88     }
89 }
View Code

搜索代码：（略丑）

  1 #include<cstring>
  2 #include<cstdlib>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<queue>
  6 #include<iostream>
  7 using namespace std;
  8 int used[105][105];//判断两个杯子装可乐的情况是否出现过
  9 struct node
 10 {
 11     int m,n,s,t;
 12 };
 13 int m,n,s;
 14 int bfs()
 15 {
 16     queue<node> q;
 17     node  now,next;
 18     now.m=now.n=now.t=0;
 19     now.s=s;
 20     q.push(now);
 21     memset(used,0,sizeof(used));
 22     used[0][0]=1;
 23     while(!q.empty())
 24     {
 25         now=q.front();
 26         q.pop();
 27         if(now.n==now.s&&now.s==s/2)
 28             return now.t;
 29         if(now.m+now.s>m)  //s->m
 30         {
 31             next.m=m,next.n=now.n,next.s=now.m+now.s-m,next.t=now.t+1;
 32             if(used[next.m][next.n]==0)
 33             {
 34                 used[next.m][next.n]=1;
 35                 q.push(next);
 36             }
 37         }
 38         else
 39         {
 40             next.m=now.m+now.s,next.n=now.n,next.s=0,next.t=now.t;
 41              if(used[next.m][next.n]==0)
 42             {
 43                 used[next.m][next.n]=1;
 44                 q.push(next);
 45             }
 46         }
 47         if(now.n+now.s>n)  //s->n
 48         {
 49             next.m=now.m,next.n=n,next.s=now.n+now.s-n,next.t=now.t+1;
 50             if(used[next.m][next.n]==0)
 51             {
 52                 used[next.m][next.n]=1;
 53                 q.push(next);
 54             }
 55         }
 56         else
 57         {
 58             next.m=now.m,next.n=now.n+now.s,next.s=0,next.t=now.t+1;
 59             if(used[next.m][next.n]==0)
 60             {
 61                 used[next.m][next.n]=1;
 62                 q.push(next);
 63             }
 64         }
 65         if(now.m+now.n>n)  //m->n
 66         {
 67             next.m=now.m+now.n-n,next.n=n,next.s=now.s,next.t=now.t+1;
 68             if(used[next.m][next.n]==0)
 69             {
 70                 used[next.m][next.n]=1;
 71                 q.push(next);
 72             }
 73         }
 74         else
 75         {
 76             next.m=0,next.n=now.n+now.m,next.s=now.s,next.t=now.t+1;
 77             if(used[next.m][next.n]==0)
 78             {
 79                 used[next.m][next.n]=1;
 80                 q.push(next);
 81             }
 82         }
 83         if(now.m+now.s>s)  //m->s
 84         {
 85             next.m=now.m+now.s-s,next.n=now.n,next.s=s,next.t=now.t+1;
 86             if(used[next.m][next.n]==0)
 87             {
 88                 used[next.m][next.n]=1;
 89                 q.push(next);
 90             }
 91         }
 92         else
 93         {
 94             next.m=0,next.n=now.n,next.s=now.m+now.s,next.t=now.t+1;
 95             if(used[next.m][next.n]==0)
 96             {
 97                 used[next.m][next.n]=1;
 98                 q.push(next);
 99             }
100         }
101         if(now.n+now.m>m)  //n->m
102         {
103             next.m=m,next.n=now.n+now.m-m,next.s=now.s,next.t=now.t+1;
104             if(used[next.m][next.n]==0)
105             {
106                 used[next.m][next.n]=1;
107                 q.push(next);
108             }
109         }
110         else
111         {
112             next.m=now.n+now.m,next.n=0,next.s=now.s,next.t=now.t+1;
113             if(used[next.m][next.n]==0)
114             {
115                 used[next.m][next.n]=1;
116                 q.push(next);
117             }
118         }
119         if(now.n+now.s>s) //n->s
120         {
121             next.m=now.m,next.n=now.n+now.s-s,next.s=s,next.t=now.t+1;
122             if(used[next.m][next.n]==0)
123             {
124                 used[next.m][next.n]=1;
125                 q.push(next);
126             }
127         }
128         else
129         {
130             next.m=now.m,next.n=0,next.s=now.n+now.s,next.t=now.t+1;
131             if(used[next.m][next.n]==0)
132             {
133                 used[next.m][next.n]=1;
134                 q.push(next);
135             }
136         }
137     }
138     return -1;
139 }
140 int main()
141 {
142     while(cin>>s>>n>>m,n+m+s)
143     {
144         if(s%2)
145             printf("NO\n");
146         else
147         {
148             if(n<m) swap(n,m);
149             int ans=bfs();
150             if(ans==-1)
151                 printf("NO\n");
152             else
153                 printf("%d\n",ans);
154         }
155     }
156     return 0;
157 }
View Code


HDU 2612

链接：http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=15717

题意：两个小伙伴打算在KFC见面。。求两人到KFC最短的路程。KFC的个数>=1

题解：感觉BFS DFS都可以做吧。。我的做法是暴力BFS了。。。要注意其中有的KFC是不能到的。

代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 char s[200+5][200+5];
 8 int vis[200+5][200+5];
 9 int tm[200+5][200+5];
10 int dir[4][2] = {1,0,-1,0,0,1,0,-1};
11 struct st{
12         int x,y;
13         int t;
14 };
15 queue<st>q;
16 void bfs(int x,int y,int n,int m)
17 {
18         memset(vis,0,sizeof(vis));
19         st f;
20         f.x = x;
21         f.y = y;
22         f.t = 0;
23         q.push(f);
24         vis[f.x][f.y] = 1;
25         while(!q.empty())
26         {
27                 f = q.front();
28                 q.pop();
29                 for( int i = 0;i < 4; i++ )
30                 {
31                         st g;
32                         g.x = f.x+dir[i][0];
33                         g.y = f.y+dir[i][1];
34                         g.t = f.t+1;
35                         if((s[g.x][g.y]=='.'||s[g.x][g.y]=='@')&&!vis[g.x][g.y]&&g.x>=1&&g.x<=n&&g.y>=1&&g.y<=m)
36                         {
37                                 q.push(g);
38                                 if(s[g.x][g.y]=='@')
39                                         tm[g.x][g.y]+=g.t;
40                                 vis[g.x][g.y] = 1;
41                         }
42                 }
43         }
44 }
45 
46 int main()
47 {
48         int n,m;
49         while(cin>>n>>m)
50         {
51                 memset(tm,0,sizeof(tm));
52                 memset(s,0,sizeof(s));
53                 int x1,x2,y1,y2;
54                 for(int i = 1; i <= n; i++ ) cin>>s[i]+1;
55                 for(int i = 1; i <= n; i++ )
56                         for(int j = 1; j <= m; j++ )
57                         {
58                                 if( s[i][j] == 'Y') {x1 = i; y1 = j;}
59                                 if( s[i][j] == 'M') {x2 = i; y2 = j;}
60                         }
61                 bfs(x1,y1,n,m);
62                 bfs(x2,y2,n,m);
63                 int mi = 999999989;
64                 for(int i = 1;i <= n; i++)
65                         for(int j = 1; j <= m;j++ )
66                         {
67                                 if(s[i][j]=='@'&&tm[i][j]!=0)
68                                 mi=min(mi,tm[i][j]);
69                         }
70                 printf("%d\n",mi*11);
71         }
72         return 0;
73 }
View Code

 

posted on 2015-08-17 21:48  小松song  阅读(147)  评论(0)    收藏  举报

导航