HDU 5372 Segment Gameツリー配列
5582 ワード
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Segment Game
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 273 Accepted Submission(s): 48
Problem Description
Lillian is a clever girl so that she has lots of fans and often receives gifts from her fans.
One day Lillian gets some segments from her fans Lawson with lengths of 1,2,3... and she intends to display them by adding them to a number line.At the i-th add operation,she will put the segment with length of i on the number line.Every time she put the segment on the line,she will count how many entire segments on that segment.During the operation ,she may delete some segments on the line.(Segments are mutually independent)
Input
There are multiple test cases.
The first line of each case contains a integer n — the number of operations(1<=n<=
2∗105 ,
∑n <=
7∗105 )
Next n lines contain the descriptions of the operatons,one operation per line.Each operation contains two integers a , b.
if a is 0,it means add operation that Lilian put a segment on the position b(|b|<
109 ) of the line.
(For the i-th add operation,she will put the segment on [b,b+i] of the line, with length of i.)
if a is 1,it means delete operation that Lilian will delete the segment which was added at the b-th add operation.
Output
For i-th case,the first line output the test case number.
Then for each add operation,ouput how many entire segments on the segment which Lillian newly adds.
Sample Input
Sample Output
Source
2015 Multi-University Training Contest 7
タイトル:
1つの線分を挿入するたびに、または既存の線分を削除し、挿入するたびに現在挿入されている線分を出力して、存在するいくつかの線分を完全に上書きできます.
問題:新しく挿入された線分について、その線分の左端より大きい線分があるかどうかを問い合せます.さらに、どのくらいのセグメントの右端点がそのセグメントの右端点より大きいかをクエリします.両者の差が答えです.2つの木の配列でできます.時間複雑度nlogn
全部で4つの場合、絵を描くと発見できるはずです.
Segment Game
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 273 Accepted Submission(s): 48
Problem Description
Lillian is a clever girl so that she has lots of fans and often receives gifts from her fans.
One day Lillian gets some segments from her fans Lawson with lengths of 1,2,3... and she intends to display them by adding them to a number line.At the i-th add operation,she will put the segment with length of i on the number line.Every time she put the segment on the line,she will count how many entire segments on that segment.During the operation ,she may delete some segments on the line.(Segments are mutually independent)
Input
There are multiple test cases.
The first line of each case contains a integer n — the number of operations(1<=n<=
2∗105 ,
∑n <=
7∗105 )
Next n lines contain the descriptions of the operatons,one operation per line.Each operation contains two integers a , b.
if a is 0,it means add operation that Lilian put a segment on the position b(|b|<
109 ) of the line.
(For the i-th add operation,she will put the segment on [b,b+i] of the line, with length of i.)
if a is 1,it means delete operation that Lilian will delete the segment which was added at the b-th add operation.
Output
For i-th case,the first line output the test case number.
Then for each add operation,ouput how many entire segments on the segment which Lillian newly adds.
Sample Input
3
0 0
0 3
0 1
5
0 1
0 0
1 1
0 1
0 0
Sample Output
Case #1:
0
0
0
Case #2:
0
1
0
2
Hint
For the second case in the sample: At the first add operation,Lillian adds a segment [1,2] on the line. At the second add operation,Lillian adds a segment [0,2] on the line. At the delete operation,Lillian deletes a segment which added at the first add operation. At the third add operation,Lillian adds a segment [1,4] on the line. At the fourth add operation,Lillian adds a segment [0,4] on the line
Source
2015 Multi-University Training Contest 7
タイトル:
1つの線分を挿入するたびに、または既存の線分を削除し、挿入するたびに現在挿入されている線分を出力して、存在するいくつかの線分を完全に上書きできます.
問題:新しく挿入された線分について、その線分の左端より大きい線分があるかどうかを問い合せます.さらに、どのくらいのセグメントの右端点がそのセグメントの右端点より大きいかをクエリします.両者の差が答えです.2つの木の配列でできます.時間複雑度nlogn
全部で4つの場合、絵を描くと発見できるはずです.
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) pt(x / 10);
putchar(x % 10 + '0');
}
typedef pair<int, int> pii;
typedef long long ll;
const int N = 450007;
struct Tree {
int c[N], maxn;
void init(int n) { maxn = n; for (int i = 0; i <= n; i++)c[i] = 0; }
int lowbit(int x) { return x&-x; }
int sum(int x) {
int ans = 0;
while (x)ans += c[x], x -= lowbit(x);
return ans;
}
void update(int pos, int val) {
while (pos <= maxn)c[pos] += val, pos += lowbit(pos);
}
}A, B;
int n;
set<pii> s;
int op[N], l[N], r[N];
pii a[N];
vector<int>G;
int main() {
int cas = 0;
while (cin>>n) {
G.clear();
int top = 0;
for (int i = 1; i <= n; i++) {
rd(op[i]), rd(l[i]);
if (op[i] == 0)
{
G.push_back(l[i]);
r[i] = l[i] + (++top);
G.push_back(r[i]);
}
}
printf("Case #%d:
", ++cas);
sort(G.begin(), G.end()); G.erase(unique(G.begin(), G.end()), G.end());
top = 0;
for (int i = 1; i <= n; i++)
if (op[i] == 0) {
l[i] = lower_bound(G.begin(), G.end(), l[i]) - G.begin() + 1;
r[i] = lower_bound(G.begin(), G.end(), r[i]) - G.begin() + 1;
a[++top] = { l[i], r[i] };
}
A.init(G.size()); B.init(G.size());
int all = 0;
for (int i = 1; i <= n; i++)
{
if (op[i] == 0)
{
int ans = B.sum(r[i]);
ans -= A.sum(l[i]-1);
pt(ans); putchar('
');
A.update(l[i], 1);
B.update(r[i], 1);
all++;
}
else {
A.update(a[l[i]].first, -1);
B.update(a[l[i]].second, -1);
all--;
}
}
}
return 0;
}
/*
99
7
1 2 2 1 3 !1 2
/*
1
????0
1 8 3 7 2
*/