HDOJ 4302 Holedox Eating (multiset || 线段树)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4302

 

思路：很多人说这题是个线段树题。。。不过当时没想出来怎么用线段树做~后面再补吧。。。

比赛时用的方法就是用multiset模拟，中间WA好几次打了几个补丁才过的（果然我好弱）。。。有一个错是走数据才看出来的，就是当当前位置比set里的所有元素都要大时，lower_bound()找不到合适的元素会返回end()的迭代器，这种情况需要特殊处理一下。

 

附个数据吧~

HDOJ 4302 数据

50



6 16

1

0 0

0 2

0 1

1

0 3

0 1

1

0 0

1

1

1

0 5

0 2

1

1







27 9

1

1

0 21

0 16

1

1

0 0

1

0 13







1 24

1

1

1

0 0

1

1

0 0

1

1

0 0

1

0 0

1

0 0

1

0 0

1

1

1

1

1

0 0

1

0 0







16 11

1

0 11

0 8

1

0 3

1

0 12

0 15

0 11

0 7

0 11







1 10

0 0

0 0

1

1

1

0 0

0 0

0 0

1

1







30 22

0 21

0 11

1

0 14

0 2

0 27

0 1

1

1

1

0 1

0 25

0 0

1

0 26

0 20

1

0 20

0 8

1

1

1







11 19

0 3

0 6

0 7

1

1

0 5

1

1

1

0 8

0 3

1

0 6

0 9

1

1

0 1

0 10

0 2







3 26

0 0

0 2

1

0 0

0 0

1

1

0 1

0 0

1

0 0

0 1

0 2

1

1

1

0 2

1

0 2

0 1

0 1

1

1

1

1

1







16 1

1







14 7

1

0 8

1

1

1

1

0 6







1 21

0 0

0 0

0 0

1

1

1

1

0 0

1

0 0

1

0 0

1

0 0

0 0

0 0

1

0 0

0 0

1

1







17 20

0 0

1

1

0 15

1

1

1

0 2

0 13

0 4

0 2

0 6

1

1

0 14

0 14

1

1

0 12

0 13







18 2

0 12

0 13







10 3

1

1

0 8







14 27

1

0 7

1

1

1

0 2

0 7

1

0 13

1

0 6

0 11

1

0 1

1

0 1

1

1

1

1

1

1

1

1

1

1

1







14 26

0 13

1

0 6

0 8

1

0 0

1

0 3

0 2

1

1

1

0 1

1

0 8

0 4

0 12

1

0 6

1

0 10

1

0 11

1

0 13

0 12







22 11

0 6

0 15

1

1

1

0 2

1

0 7

1

1

0 21







3 11

0 2

1

0 0

0 0

0 0

1

1

1

0 2

1

1







29 24

0 13

1

1

1

1

1

0 3

1

0 7

1

0 27

1

0 11

1

1

1

1

1

1

0 28

1

0 24

0 5

1







20 0







8 22

1

1

1

1

1

1

0 1

0 3

0 2

1

1

0 3

0 0

1

0 5

1

1

1

0 3

1

0 7

1







11 2

1

1







21 10

1

1

0 17

0 6

0 2

1

1

0 7

1

1







7 14

1

0 1

1

0 2

1

1

1

1

1

1

1

1

1

0 0







21 28

0 13

1

0 7

0 19

1

1

1

0 14

1

0 6

0 14

1

0 20

0 18

0 11

1

0 7

1

0 9

1

0 19

1

0 12

0 20

1

1

1

1







18 18

0 3

0 9

0 10

0 13

0 14

0 1

1

0 11

0 10

1

0 14

1

0 16

0 12

1

0 3

0 2

1







9 23

0 4

0 5

1

0 1

0 0

0 7

1

1

1

1

0 7

0 8

0 8

1

0 0

1

1

1

0 0

1

1

0 3

1







4 1

1







30 4

0 6

1

0 29

0 18







25 24

1

0 1

0 13

0 4

0 1

1

0 11

0 19

1

0 9

1

1

0 1

0 4

0 23

0 8

0 6

1

1

1

0 14

1

1

1







23 18

0 4

1

0 1

0 16

0 7

1

1

0 8

0 4

0 5

1

1

0 4

1

0 22

0 20

0 14

1







29 25

1

1

0 13

0 18

1

1

0 20

1

0 20

0 7

1

1

1

0 16

0 21

1

0 7

0 25

1

1

0 24

0 19

1

0 1

0 9







8 10

0 1

1

1

0 2

1

0 1

0 7

0 1

0 3

0 6







11 19

0 8

1

1

0 5

0 6

1

0 0

0 4

0 4

0 9

1

1

1

1

0 1

0 10

1

1

1







21 2

1

0 6







24 27

1

1

0 11

1

0 0

0 14

0 4

0 14

0 18

1

1

1

0 0

0 9

0 12

0 19

1

0 16

0 6

0 10

0 13

1

0 16

0 4

0 2

1

0 10







9 29

1

0 5

1

0 0

0 1

0 8

1

0 4

0 1

0 3

1

0 0

0 4

0 8

0 7

0 4

1

1

0 0

0 0

1

1

1

0 6

0 4

1

1

0 4

0 6







18 15

1

1

0 10

0 10

1

1

0 14

0 16

0 17

0 1

1

1

0 15

1

0 12







17 11

0 5

0 7

1

0 0

1

0 11

0 8

1

1

1

1







13 24

1

0 8

0 0

0 11

1

1

0 10

0 3

0 4

1

0 9

0 5

1

1

0 5

0 8

1

1

0 11

0 7

0 11

1

0 5

0 9







23 25

1

0 3

1

1

0 6

0 13

1

1

0 9

1

1

0 8

1

0 18

1

0 15

0 0

0 10

0 21

0 17

1

0 13

1

1

0 3







9 1

0 3







8 0







20 1

1







1 13

0 0

0 0

0 0

1

1

0 0

1

1

1

0 0

0 0

0 0

0 0







11 6

0 5

1

1

1

1

0 10







16 6

1

1

1

0 1

1

1







3 12

1

1

0 0

0 2

0 2

0 0

1

0 0

0 0

1

0 0

0 2







29 20

0 11

1

1

1

0 28

0 2

1

0 7

0 23

1

0 28

0 23

1

1

0 8

0 12

0 27

0 17

1

0 28







27 1

0 26 输出： Case 1: 6 Case 2: 42 Case 3: 0 Case 4: 11 Case 5: 0 Case 6: 48 Case 7: 16 Case 8: 3 Case 9: 0 Case 10: 8 Case 11: 0 Case 12: 28 Case 13: 0 Case 14: 0 Case 15: 35 Case 16: 36 Case 17: 33 Case 18: 6 Case 19: 84 Case 20: 0 Case 21: 17 Case 22: 0 Case 23: 17 Case 24: 2 Case 25: 60 Case 26: 10 Case 27: 33 Case 28: 0 Case 29: 6 Case 30: 29 Case 31: 22 Case 32: 52 Case 33: 2 Case 34: 26 Case 35: 0 Case 36: 22 Case 37: 16 Case 38: 17 Case 39: 22 Case 40: 17 Case 41: 33 Case 42: 0 Case 43: 0 Case 44: 0 Case 45: 0 Case 46: 5 Case 47: 1 Case 48: 0 Case 49: 45 Case 50: 0

 

multiset 解法（打了好多补丁，好丑。。。）

 

6771936

2012-09-15 16:45:47

Accepted

4302

171MS

280K

2665 B

C++

AbandonZHANG

 

View Code

 

线段树解法（用线段树求[0,now]区间最大数，求[now,l]区间最小的数）

 

6774072

2012-09-15 22:13:02

Accepted

4302

296MS

1808K

3792 B

C++

AbandonZHANG

 

View Code

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstdlib>

  4 #include <cmath>

  5 #include <iomanip>

  6 #include <climits>

  7 #include <vector>

  8 #include <stack>

  9 #include <queue>

 10 #include <set>

 11 #include <map>

 12 #include <algorithm>

 13 #include <string>

 14 #include <cstring>

 15 #define MID(x,y) ( ( x + y ) >> 1 )

 16 #define FOR(i,s,t) for(int i=s; i<t; i++)

 17 

 18 using namespace std;

 19 

 20 typedef long long LL;

 21 

 22 const int N=100005;

 23 int M;

 24 int sum[N<<2];

 25 

 26 void PushUp(int rt)

 27 {

 28     sum[rt]=sum[rt<<1]+sum[rt<<1|1];

 29 }

 30 

 31 void BuildTree(int n)

 32 {

 33     for (M=1;M<=n+2;M<<=1);

 34     for (int i=1;i<N<<2;i++)

 35         sum[i]=0;

 36 }

 37 

 38 void Update(int n,int v)

 39 {

 40     for (sum[n+=M]+=v,n>>=1;n;n>>=1)

 41         PushUp(n);

 42 }

 43 

 44 int Qsum(int s,int t,int l,int r,int rt)

 45 {

 46     if (s<=l && r<=t)

 47     {

 48         return sum[rt];

 49     }

 50 

 51     int mid=MID(l,r);

 52     int ans=0;

 53     if (s<=mid) ans+=Qsum(s,t,l,mid,rt<<1);

 54     if (mid<t)  ans+=Qsum(s,t,mid+1,r,rt<<1|1);

 55     return ans;

 56 }

 57 

 58 int Querymax(int s,int l,int r,int rt)

 59 {

 60     if (l==r)

 61     {

 62         return l;

 63     }

 64 

 65     int mid=MID(l,r);

 66     if (sum[rt<<1]>=s)

 67     {

 68         return Querymax(s,l,mid,rt<<1);

 69     }

 70     else return Querymax(s-sum[rt<<1],mid+1,r,rt<<1|1);

 71 }

 72 

 73 int Querymin(int s,int l,int r,int rt)

 74 {

 75     if (l==r)

 76     {

 77         return l;

 78     }

 79 

 80     int mid=MID(l,r);

 81     if (sum[rt<<1|1]>=s)

 82     {

 83         return Querymin(s,mid+1,r,rt<<1|1);

 84     }

 85     else return Querymin(s-sum[rt<<1|1],l,mid,rt<<1);

 86 }

 87 

 88 int main()

 89 {

 90     //freopen("test.in","r+",stdin);

 91 

 92     int t,caseo=1;

 93     scanf("%d",&t);

 94     while(t--)

 95     {

 96         printf("Case %d: ",caseo++);

 97         int ans=0;

 98         int l,n;

 99         scanf("%d%d",&l,&n);

100         BuildTree(l);

101         int h=0;

102         int dir=0;

103         for (int i=0;i<n;i++)

104         {

105             int p;

106             scanf("%d",&p);

107             if (p==0)

108             {

109                 int num;

110                 scanf("%d",&num);

111                 Update(num,1);

112             }

113             else

114             {

115                 if (sum[1]==0)  continue;

116                 int a1=Qsum(0,h,0,M-1,1);

117                 int a2=Qsum(h,M,0,M-1,1);

118                 if (a1==0)

119                 {

120                     int b2=Querymin(a2,0,M-1,1);

121                     dir=1;

122                     ans+=b2-h;

123                     h=b2;

124                     Update(b2,-1);

125                 }

126                 else if (a2==0)

127                 {

128                     int b1=Querymax(a1,0,M-1,1);

129                     dir=-1;

130                     ans+=h-b1;

131                     h=b1;

132                     Update(b1,-1);

133                 }

134                 else

135                 {

136                     int b1=Querymax(a1,0,M-1,1);

137                     int b2=Querymin(a2,0,M-1,1);

138                     if (b2-h>h-b1)

139                     {

140                         ans+=h-b1;

141                         h=b1;

142                         dir=-1;

143                         Update(b1,-1);

144                     }

145                     else if (b2-h<h-b1)

146                     {

147                         ans+=b2-h;

148                         h=b2;

149                         dir=1;

150                         Update(b2,-1);

151                     }

152                     else

153                     {

154                         ans+=b2-h;

155                         if (dir==1)

156                         {

157                             h=b2;

158                             Update(b2,-1);

159                         }

160                         else

161                         {

162                             h=b1;

163                             Update(b1,-1);

164                         }

165                     }

166                 }

167             }

168         }

169         printf("%d\n",ans);

170     }

171     return 0;

172 }