bzoj1588: [HNOI2002]营业额统计
地址:http://www.lydsy.com/JudgeOnline/problem.php?id=1588
题目:
1588: [HNOI2002]营业额统计
Time Limit: 5 Sec Memory Limit: 162 MBSubmit: 17267 Solved: 7013
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Description
营业额统计 Tiger最近被公司升任为营业部经理,他上任后接受公司交给的第一项任务便是统计并分析公司成立以来的营业情况。 Tiger拿出了公司的账本,账本上记录了公司成立以来每天的营业额。分析营业情况是一项相当复杂的工作。由于节假日,大减价或者是其他情况的时候,营业额会出现一定的波动,当然一定的波动是能够接受的,但是在某些时候营业额突变得很高或是很低,这就证明公司此时的经营状况出现了问题。经济管理学上定义了一种最小波动值来衡量这种情况: 该天的最小波动值 当最小波动值越大时,就说明营业情况越不稳定。 而分析整个公司的从成立到现在营业情况是否稳定,只需要把每一天的最小波动值加起来就可以了。你的任务就是编写一个程序帮助Tiger来计算这一个值。 第一天的最小波动值为第一天的营业额。 输入输出要求
Input
第一行为正整数 ,表示该公司从成立一直到现在的天数,接下来的n行每行有一个整数(有可能有负数) ,表示第i
天公司的营业额。
天数n<=32767,
每天的营业额ai <= 1,000,000。
最后结果T<=2^31
Output
输出文件仅有一个正整数,即Sigma(每天最小的波动值) 。结果小于2^31 。
Sample Input
6
5
1
2
5
4
6
5
1
2
5
4
6
Sample Output
12
HINT
结果说明:5+|1-5|+|2-1|+|5-5|+|4-5|+|6-5|=5+4+1+0+1+1=12
该题数据bug已修复.----2016.5.15
思路:
复习splay
1 /************************************************************** 2 Problem: 1588 3 User: weeping 4 Language: C++ 5 Result: Accepted 6 Time:196 ms 7 Memory:4028 kb 8 ****************************************************************/ 9 10 #include <bits/stdc++.h> 11 12 using namespace std; 13 14 #define lc ch[x][0] 15 #define rc ch[x][1] 16 17 struct SplayTree 18 { 19 20 const static int maxn = 1e5 + 15; 21 22 int tot,root,ch[maxn][2], key[maxn], val[maxn], sz[maxn], rev[maxn], fa[maxn]; 23 24 inline void init( int x, int ky, int v = 0, int par = 0 ) 25 { 26 lc=rc=0, fa[x]= par, key[x] = ky, val[x] = v, sz[x] = 1, rev[x] = 0; 27 } 28 29 inline void init() 30 { 31 init( 0, 0, 0 ); 32 sz[0] = 0; 33 tot = root = 0 ; 34 } 35 36 inline void push_up(int x) 37 { 38 sz[x] = sz[lc] + sz[rc] + 1; 39 } 40 41 inline void reverse(int x) 42 { 43 rev[x] ^= 1, swap( lc, rc); 44 } 45 46 inline void push_down(int x) 47 { 48 if(rev[x]) 49 { 50 if(lc) reverse(lc); 51 if(rc) reverse(rc); 52 rev[x] = 0; 53 } 54 } 55 56 void rotate( int x) 57 { 58 int f = fa[x], gf = fa[f]; 59 int t1 = (ch[f][1] == x), t2 = (ch[gf][1] == f); 60 if( gf ) ch[gf][t2] = x; 61 fa[x] = gf, ch[f][t1] = ch[x][1^t1], fa[ch[f][t1]] = f; 62 ch[x][t1^1] = f, fa[f] = x; 63 push_up( f ), push_up( x ); 64 } 65 66 void splay( int x, int tar ) 67 { 68 for(int f = fa[x], gf = fa[f]; f != tar; rotate(x), f = fa[x], gf = fa[f]) 69 if(gf != tar) 70 rotate( ((ch[f][1] == x) == (ch[gf][1] == f) )? f: x); 71 if( !tar ) root = x; 72 } 73 74 void insert( int ky, int v) 75 { 76 int x = root, ls = root; 77 while(x) 78 { 79 push_down(x); 80 sz[x] ++, ls = x; 81 x = ch[x][ky > key[x]]; 82 } 83 init( ++tot, ky, v, ls); 84 ch[ls][ky > key[ls]] = tot; 85 splay( tot, 0); 86 } 87 88 int find( int ky) 89 { 90 int x = root; 91 while(x) 92 { 93 push_down(x); 94 if(key[x] == ky) break; 95 x = ch[x][ky > key[x]]; 96 } 97 if(x) splay(x,0); 98 else x = -1; 99 return x; 100 } 101 102 // Delete Root 103 void Delete() 104 { 105 if( !ch[root][0] ) 106 { 107 fa[ ch[root][1] ] = 0 ; 108 root = ch[root][1]; 109 } 110 else 111 { 112 int cur = ch[root][0]; 113 while( ch[cur][1] ) cur = ch[cur][1]; 114 splay( cur, root ); 115 ch[cur][1] = ch[root][1]; 116 root = cur, fa[cur] = 0, fa[ch[root][1]] = root; 117 push_up( root ); 118 } 119 } 120 121 int kth( int k) 122 { 123 int x = root; 124 if(sz[x] < k) return -1; 125 while(x) 126 { 127 push_down(x); 128 if(k == sz[lc] + 1) break; 129 if(k > sz[lc]) 130 k -= sz[lc] + 1, x = rc; 131 else 132 x = lc; 133 } 134 if(x) splay(x,0); 135 else x = -1; 136 return x; 137 } 138 139 int pred( void) 140 { 141 int x = root; 142 if(!x || !lc) return -1; 143 x = lc; 144 while(rc) push_down(x), x = rc; 145 //splay( x, 0); 146 return x; 147 } 148 149 int succ( void) 150 { 151 int x = root; 152 if(!x || !rc) return -1; 153 x = rc; 154 while(lc) push_down(x), x = lc; 155 //splay( x, 0); 156 return x; 157 } 158 159 void debug( int x ) 160 { 161 if( !x ) return; 162 if(lc) debug( lc ); 163 printf("%d ", key[x] ); 164 if(rc) debug( rc ); 165 } 166 167 int go(int x) 168 { 169 insert(x,0); 170 int ans = 0x3f3f3f3f, ta = pred(), tb = succ(); 171 if(ta!=-1) 172 ans = abs(key[ta] - x); 173 if(tb!=-1) 174 ans = min( ans, abs(key[tb] - x)); 175 //printf("==%d\n",ans); 176 return ans; 177 } 178 } sp; 179 180 int main(void) 181 { 182 sp.init(); 183 int n,ls,ans=0; 184 scanf("%d%d",&n,&ls); 185 sp.insert(ls,0); 186 ans=ls; 187 for(int i=1;i<n;i++) 188 scanf("%d",&ls),ans+=sp.go(ls); 189 printf("%d\n",ans); 190 return 0; 191 }
/**************************************************************
Problem: 1588
User: weeping
Language: C++
Result: Accepted
Time:196 ms
Memory:4028 kb
****************************************************************/
#include <bits/stdc++.h>
using
namespace
std;
#define lc ch[x][0]
#define rc ch[x][1]
struct
SplayTree
{
const
static
int
maxn = 1e5 + 15;
int
tot,root,ch[maxn][2], key[maxn], val[maxn], sz[maxn], rev[maxn], fa[maxn];
inline
void
init(
int
x,
int
ky,
int
v = 0,
int
par = 0 )
{
lc=rc=0, fa[x]= par, key[x] = ky, val[x] = v, sz[x] = 1, rev[x] = 0;
}
inline
void
init()
{
init( 0, 0, 0 );
sz[0] = 0;
tot = root = 0 ;
}
inline
void
push_up(
int
x)
{
sz[x] = sz[lc] + sz[rc] + 1;
}
inline
void
reverse(
int
x)
{
rev[x] ^= 1, swap( lc, rc);
}
inline
void
push_down(
int
x)
{
if
(rev[x])
{
if
(lc) reverse(lc);
if
(rc) reverse(rc);
rev[x] = 0;
}
}
void
rotate(
int
x)
{
int
f = fa[x], gf = fa[f];
int
t1 = (ch[f][1] == x), t2 = (ch[gf][1] == f);
if
( gf ) ch[gf][t2] = x;
fa[x] = gf, ch[f][t1] = ch[x][1^t1], fa[ch[f][t1]] = f;
ch[x][t1^1] = f, fa[f] = x;
push_up( f ), push_up( x );
}
void
splay(
int
x,
int
tar )
{
for
(
int
f = fa[x], gf = fa[f]; f != tar; rotate(x), f = fa[x], gf = fa[f])
if
(gf != tar)
rotate( ((ch[f][1] == x) == (ch[gf][1] == f) )? f: x);
if
( !tar ) root = x;
}
void
insert(
int
ky,
int
v)
{
int
x = root, ls = root;
while
(x)
{
push_down(x);
sz[x] ++, ls = x;
x = ch[x][ky > key[x]];
}
init( ++tot, ky, v, ls);
ch[ls][ky > key[ls]] = tot;
splay( tot, 0);
}
int
find(
int
ky)
{
int
x = root;
while
(x)
{
push_down(x);
if
(key[x] == ky)
break
;
x = ch[x][ky > key[x]];
}
if
(x) splay(x,0);
else
x = -1;
return
x;
}
// Delete Root
void
Delete()
{
if
( !ch[root][0] )
{
fa[ ch[root][1] ] = 0 ;
root = ch[root][1];
}
else
{
int
cur = ch[root][0];
while
( ch[cur][1] ) cur = ch[cur][1];
splay( cur, root );
ch[cur][1] = ch[root][1];
root = cur, fa[cur] = 0, fa[ch[root][1]] = root;
push_up( root );
}
}
int
kth(
int
k)
{
int
x = root;
if
(sz[x] < k)
return
-1;
while
(x)
{
push_down(x);
if
(k == sz[lc] + 1)
break
;
if
(k > sz[lc])
k -= sz[lc] + 1, x = rc;
else
x = lc;
}
if
(x) splay(x,0);
else
x = -1;
return
x;
}
int
pred(
void
)
{
int
x = root;
if
(!x || !lc)
return
-1;
x = lc;
while
(rc) push_down(x), x = rc;
//splay( x, 0);
return
x;
}
int
succ(
void
)
{
int
x = root;
if
(!x || !rc)
return
-1;
x = rc;
while
(lc) push_down(x), x = lc;
//splay( x, 0);
return
x;
}
void
debug(
int
x )
{
if
( !x )
return
;
if
(lc) debug( lc );
printf
(
"%d "
, key[x] );
if
(rc) debug( rc );
}
int
go(
int
x)
{
insert(x,0);
int
ans = 0x3f3f3f3f, ta = pred(), tb = succ();
if
(ta!=-1)
ans =
abs
(key[ta] - x);
if
(tb!=-1)
ans = min( ans,
abs
(key[tb] - x));
//printf("==%d\n",ans);
return
ans;
}
} sp;
int
main(
void
)
{
sp.init();
int
n,ls,ans=0;
scanf
(
"%d%d"
,&n,&ls);
sp.insert(ls,0);
ans=ls;
for
(
int
i=1;i<n;i++)
scanf
(
"%d"
,&ls),ans+=sp.go(ls);
printf
(
"%d\n"
,ans);
return
0;
}
作者:weeping
出处:www.cnblogs.com/weeping/
本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。