ZOJ - 3961模拟题中如何处理区间问题的长尾词疑问？
- 内容介绍
- 文章标签
- 相关推荐
本文共计1002个文字，预计阅读时间需要5分钟。
ACM（ACMers的聊天信使）是一款由Marjar科技公司开发的知名即时通讯软件。为了吸引更多用户，Marjar公司老板Edward最近为该软件添加了新功能。这个新功能是……
ACM (ACMers’ Chatting Messenger) is a famous instant messaging software developed by Marjar Technology Company. To attract more users, Edward, the boss of Marjar Company, has recently added a new feature to the software. The new feature can be described as follows:
If two users, A and B, have been sending messages to each other on the last m consecutive days, the “friendship point” between them will be increased by 1 point.
More formally, if user A sent messages to user B on each day between the (i - m + 1)-th day and the i-th day (both inclusive), and user B also sent messages to user A on each day between the (i - m + 1)-th day and the i-th day (also both inclusive), the “friendship point” between A and B will be increased by 1 at the end of the i-th day.
Given the chatting logs of two users A and B during n consecutive days, what’s the number of the friendship points between them at the end of the n-th day (given that the initial friendship point between them is 0)?
Input
There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10), indicating the number of test cases. For each test case:
The first line contains 4 integers n (1 ≤ n ≤ 109), m (1 ≤ m ≤ n), x and y (1 ≤ x, y ≤ 100). The meanings of n and m are described above, while x indicates the number of chatting logs about the messages sent by A to B, and y indicates the number of chatting logs about the messages sent by B to A.
For the following x lines, the i-th line contains 2 integers la, i and ra, i (1 ≤ la, i ≤ ra, i ≤ n), indicating that A sent messages to B on each day between the la, i-th day and the ra, i-th day (both inclusive).
For the following y lines, the i-th line contains 2 integers lb, i and rb, i (1 ≤ lb, i ≤ rb, i ≤ n), indicating that B sent messages to A on each day between the lb, i-th day and the rb, i-th day (both inclusive).
It is guaranteed that for all 1 ≤ i < x, ra, i + 1 < la, i + 1 and for all 1 ≤ i < y, rb, i + 1 < lb, i + 1.
Output
For each test case, output one line containing one integer, indicating the number of friendship points between A and B at the end of the n-th day.
Sample Input
2
10 3 3 2
1 3
5 8
10 10
1 8
10 10
5 3 1 1
1 2
4 5
Sample Output
3
0
Hint
For the first test case, user A and user B send messages to each other on the 1st, 2nd, 3rd, 5th, 6th, 7th, 8th and 10th day. As m = 3, the friendship points between them will be increased by 1 at the end of the 3rd, 7th and 8th day. So the answer is 3.
题意：询问重叠区间长度满足条件的数量；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
struct cht {
int l, r;
}a[maxn],b[maxn];
int n, m;
int main()
{
ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--) {
cin >> n >> m;
int x, y;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
cin >> x >> y;
for (int i = 1; i <= x; i++)cin >> a[i].l >> a[i].r;
for (int i = 1; i <= y; i++)cin >> b[i].l >> b[i].r;
int cnt = 0;
for (int i = 1; i <= x; i++) {
if (a[i].r - a[i].l+1 < m)continue;
for (int j = 1; j <= y; j++) {
if (b[j].r - b[j].l+1 < m)continue;
int maxx = max(a[i].l, b[j].l);
int minn = min(a[i].r, b[j].r);
int len = minn - maxx + 1;
if (len >= m) {
cnt += (len - m + 1);
}
}
}
cout << cnt << endl;
}
}
本文共计1002个文字，预计阅读时间需要5分钟。
ACM（ACMers的聊天信使）是一款由Marjar科技公司开发的知名即时通讯软件。为了吸引更多用户，Marjar公司老板Edward最近为该软件添加了新功能。这个新功能是……
ACM (ACMers’ Chatting Messenger) is a famous instant messaging software developed by Marjar Technology Company. To attract more users, Edward, the boss of Marjar Company, has recently added a new feature to the software. The new feature can be described as follows:
If two users, A and B, have been sending messages to each other on the last m consecutive days, the “friendship point” between them will be increased by 1 point.
More formally, if user A sent messages to user B on each day between the (i - m + 1)-th day and the i-th day (both inclusive), and user B also sent messages to user A on each day between the (i - m + 1)-th day and the i-th day (also both inclusive), the “friendship point” between A and B will be increased by 1 at the end of the i-th day.
Given the chatting logs of two users A and B during n consecutive days, what’s the number of the friendship points between them at the end of the n-th day (given that the initial friendship point between them is 0)?
Input
There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10), indicating the number of test cases. For each test case:
The first line contains 4 integers n (1 ≤ n ≤ 109), m (1 ≤ m ≤ n), x and y (1 ≤ x, y ≤ 100). The meanings of n and m are described above, while x indicates the number of chatting logs about the messages sent by A to B, and y indicates the number of chatting logs about the messages sent by B to A.
For the following x lines, the i-th line contains 2 integers la, i and ra, i (1 ≤ la, i ≤ ra, i ≤ n), indicating that A sent messages to B on each day between the la, i-th day and the ra, i-th day (both inclusive).
For the following y lines, the i-th line contains 2 integers lb, i and rb, i (1 ≤ lb, i ≤ rb, i ≤ n), indicating that B sent messages to A on each day between the lb, i-th day and the rb, i-th day (both inclusive).
It is guaranteed that for all 1 ≤ i < x, ra, i + 1 < la, i + 1 and for all 1 ≤ i < y, rb, i + 1 < lb, i + 1.
Output
For each test case, output one line containing one integer, indicating the number of friendship points between A and B at the end of the n-th day.
Sample Input
2
10 3 3 2
1 3
5 8
10 10
1 8
10 10
5 3 1 1
1 2
4 5
Sample Output
3
0
Hint
For the first test case, user A and user B send messages to each other on the 1st, 2nd, 3rd, 5th, 6th, 7th, 8th and 10th day. As m = 3, the friendship points between them will be increased by 1 at the end of the 3rd, 7th and 8th day. So the answer is 3.
题意：询问重叠区间长度满足条件的数量；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
struct cht {
int l, r;
}a[maxn],b[maxn];
int n, m;
int main()
{
ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--) {
cin >> n >> m;
int x, y;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
cin >> x >> y;
for (int i = 1; i <= x; i++)cin >> a[i].l >> a[i].r;
for (int i = 1; i <= y; i++)cin >> b[i].l >> b[i].r;
int cnt = 0;
for (int i = 1; i <= x; i++) {
if (a[i].r - a[i].l+1 < m)continue;
for (int j = 1; j <= y; j++) {
if (b[j].r - b[j].l+1 < m)continue;
int maxx = max(a[i].l, b[j].l);
int minn = min(a[i].r, b[j].r);
int len = minn - maxx + 1;
if (len >= m) {
cnt += (len - m + 1);
}
}
}
cout << cnt << endl;
}
}