The Happy Worm(poj1974)

The Happy Worm

Time Limit: 5000MS Memory Limit: 131072K

Description

The Happy Worm lives in an m*n rectangular field. There are k stones placed in certain locations of the field. (Each square of the field is either empty, or contains a stone.) Whenever the worm sleeps, it lies either horizontally or vertically, and stretches so that its length increases as much as possible. The worm will not go in a square with a stone or out of the field. The happy worm can not be shorter than 2 squares. 

The question you are to answer is how many different positions this worm could be in while sleeping. 

Input

The first line of the input contains a single integer t (1 <= t <= 11), the number of test cases, followed by the input data for each test case. The first line of each test case contains three integers m, n, and k (1 <= m,n,k <= 131072). The input for this test case will be followed by k lines. Each line contains two integers which specify the row and column of a stone. No stone will be given twice.

Output

There should be one line per test case containing the number of positions the happy worm can be in.

Sample Input

15 5 61 52 32 44 2 4 35 1

Sample Output

9

#include<cstdio>#include<cstdlib>struct Lattice{int x, y;}la[200000];int cmpx(const void* c, const void* d){Lattice* a = (Lattice*)c;Lattice* b = (Lattice*)d;if (a->x == b->x) return a->y - b->y;else return a->x - b->x;}int cmpy(const void* c, const void* d){Lattice* a = (Lattice*)c;Lattice* b = (Lattice*)d;if (a->y == b->y) return a->x - b->x;else return a->y - b->y;}int main(){int T;scanf("%d", &T);while(T--){int m, n , k;scanf("%d%d%d", &m, &n, &k);for (int i = 0; i < k; ++i)scanf("%d%d", &la[i].x, &la[i].y);int po = 0;if (k == 0) po = m + n;else{qsort(la, k, sizeof(Lattice), cmpx);po += la[0].x - 1;if (la[0].y > 2) ++po;for (int i = 0; i < k - 1; ++i)if (la[i].x == la[i+1].x){if (la[i].y + 2 < la[i+1].y) ++po;}else{po += la[i+1].x - la[i].x - 1;if (la[i].y + 1 < n) ++po;if (la[i+1].y > 2) ++po;}if (la[k-1].y + 1 < n) ++po;po += m - la[k-1].x;qsort(la, k, sizeof(Lattice), cmpy);po += la[0].y - 1;if (la[0].x > 2) ++po;for (int i = 0; i < k - 1; ++i)if (la[i].y == la[i+1].y ){if (la[i].x + 2 < la[i+1].x) ++po;}else{po += la[i+1].y - la[i].y - 1;if (la[i].x + 1 < m) ++po;if (la[i+1].x > 2) ++po;}if (la[k-1].x + 1 < m) ++po;po += n - la[k-1].y;}printf("%d\n", po);}return 0;}