A. Orchestra
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Paul is at the orchestra. The string section is arranged in an r × c rectangular grid and is filled with violinists with the exception of n violists. Paul really likes violas, so he would like to take a picture including at least k of them. Paul can take a picture of any axis-parallel rectangle in the orchestra. Count the number of possible pictures that Paul can take.
Two pictures are considered to be different if the coordinates of corresponding rectangles are different.
Input
The first line of input contains four space-separated integers r, c, n, k (1 ≤ r, c, n ≤ 10, 1 ≤ k ≤ n) — the number of rows and columns of the string section, the total number of violas, and the minimum number of violas Paul would like in his photograph, respectively.
The next n lines each contain two integers xi and yi (1 ≤ xi ≤ r, 1 ≤ yi ≤ c): the position of the i-th viola. It is guaranteed that no location appears more than once in the input.
Output
Print a single integer — the number of photographs Paul can take which include at least k violas.
Examples
Input
2 2 1 1
1 2
Output
4
Input
3 2 3 3
1 1
3 1
2 2
Output
1
Input
3 2 3 2
1 1
3 1
2 2
Output
4
Note
We will use ‘*’ to denote violinists and ‘#’ to denote violists.
In the first sample, the orchestra looks as follows
*#
**
Paul can take a photograph of just the viola, the 1 × 2 column containing the viola, the 2 × 1 row containing the viola, or the entire string section, for 4 pictures total.
In the second sample, the orchestra looks as follows
*
*#
*
Paul must take a photograph of the entire section.
In the third sample, the orchestra looks the same as in the second sample.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
typedef long long int llint;
#define mem(a) memset(a, 0, sizeof(a))
#define pi acos(-1)
const llint maxn = 1e5+100;
/*-------------------模板-----------------------------------------*/
template <typename T>
int com(const T& v1, const T& v2) {
if (v1 < v2) return -1;
else if (v1 > v2) return 1;
return 0;
}
int a[20][20];
int b[maxn], vis[maxn];
int check(int x,int x1,int y,int y1)
{
int ans = 0;
for (int i = x;i<=x1;i++)
for (int j = y;j<=y1;j++)
if (a[i][j])
ans++;
return ans;
}
int main() {
int r, c, n, k; cin >> r >> c >> n >> k;
mem(a);
while (n --) {
int x, y; cin >> x >> y;
a[x][y] = 1;
}
// cout << k << endl;
// cout << a[1][1] << a[1][2] <<a[3][1] << endl;
int num = 0, num2 = 0, res = 0;
for (int i = 1; i<=r; i++) {
for (int ii = i; ii<=r; ii++) {
for (int j = 1; j<=c; j++) {
for (int jj = j; jj<=c; jj++) {
if (check(i, ii, j, jj) >= k) res ++ ;
}
}
}
}
cout << res << endl;
return 0;
}B. Island Puzzle
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
A remote island chain contains n islands, labeled 1 through n. Bidirectional bridges connect the islands to form a simple cycle — a bridge connects islands 1 and 2, islands 2 and 3, and so on, and additionally a bridge connects islands n and 1. The center of each island contains an identical pedestal, and all but one of the islands has a fragile, uniquely colored statue currently held on the pedestal. The remaining island holds only an empty pedestal.
The islanders want to rearrange the statues in a new order. To do this, they repeat the following process: First, they choose an island directly adjacent to the island containing an empty pedestal. Then, they painstakingly carry the statue on this island across the adjoining bridge and place it on the empty pedestal.
Determine if it is possible for the islanders to arrange the statues in the desired order.
Input
The first line contains a single integer n (2 ≤ n ≤ 200 000) — the total number of islands.
The second line contains n space-separated integers ai (0 ≤ ai ≤ n - 1) — the statue currently placed on the i-th island. If ai = 0, then the island has no statue. It is guaranteed that the ai are distinct.
The third line contains n space-separated integers bi (0 ≤ bi ≤ n - 1) — the desired statues of the ith island. Once again, bi = 0 indicates the island desires no statue. It is guaranteed that the bi are distinct.
Output
Print “YES” (without quotes) if the rearrangement can be done in the existing network, and “NO” otherwise.
Examples
Input
3
1 0 2
2 0 1
Output
YES
Input
2
1 0
0 1
Output
YES
Input
4
1 2 3 0
0 3 2 1
Output
NO
Note
In the first sample, the islanders can first move statue 1 from island 1 to island 2, then move statue 2 from island 3 to island 1, and finally move statue 1 from island 2 to island 3.
In the second sample, the islanders can simply move statue 1 from island 1 to island 2.
In the third sample, no sequence of movements results in the desired position.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
typedef long long int llint;
#define mem(a) memset(a, 0, sizeof(a))
#define pi acos(-1)
const llint maxn = 2e5+100;
/*-------------------模板-----------------------------------------*/
template <typename T>
int com(const T& v1, const T& v2) {
if (v1 < v2) return -1;
else if (v1 > v2) return 1;
return 0;
}
int a[maxn], b[maxn];
int main() {
int n; scanf("%d",&n);
for (int i = 1; i<=n; i++) scanf("%d",&a[i]);
for (int i = 1; i<=n; i++) scanf("%d",&b[i]);
int pos, f = a[1];
if (a[1] == 0) f = a[2];
for (int i = 1; i<=n; i++) {
if (f == b[i]) {
pos = i;
break;
}
}
int flag = 0;
for (int i = 1; i<=n; i++) {
if (a[i] == b[pos]) {
pos ++ ;
if (pos == n+1) pos = 1;
}
else if (a[i] == 0 && b[pos] == 0) {
pos ++ ;
if (pos == n+1) pos = 1;
}
else if (a[i] == 0 && b[pos] != 0) continue;
else if (a[i] != 0 && b[pos] == 0) {
pos ++;
if (pos == n+1) pos = 1;
if (a[i] != b[pos]) flag = 1;
pos ++ ;
if (pos == n+1) pos = 1;
}
else if (a[i] != b[pos]) {
flag = 1;
pos ++ ;
if (pos == n+1) pos = 1;
}
}
if (flag) printf("NO
");
else printf("YES
");
return 0;
}