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 }