POJ 3067 Japan

 

apan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8376		Accepted: 2235

Description

Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at one location. Write a program that calculates the number of the crossings between superhighways.

Input

The input file starts with T - the number of test cases. Each test case starts with three numbers – N, M, K. Each of the next K lines contains two numbers – the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one is the number of the city of the West coast.

Output

For each test case write one line on the standard output: 
Test case (case number): (number of crossings)

Sample Input

1
3 4 4
1 4
2 3
3 2
3 1

Sample Output

Test case 1: 5

Source

Southeastern Europe 2006

 

/* http://acm.pku.edu.cn/JudgeOnline/problem?id=3067 这题一开始使用二维树状数组,case结果正确却TLE. 后来分析了一下发现可以先对输入按照x坐标排序, 然后只需要按照y坐标来处理,这样就从二维树状数 组简化成了一维树状数组,再提交就AC了. 两个方案的代码都有,而且感觉二维树状数组也是对的 注：刚好在城市上的交点不计入最后结果 另外，交点数很大，需要用__int64，一开始用int就WA */ /* 采用二维线短树 #include <iostream> #define MAX_N 1000 using namespace std; __int64 countv[MAX_N + 1][MAX_N + 1]; int numN, numM, numK, caseNum; __int64 total; int lowbit(int p) { return p & (p ^ (p - 1)); } void modify(int row, int col, __int64 val) { int i, j; for(i = row; i <= numN; i += lowbit(i)) for(j = col; j <= numM; j += lowbit(j)) countv[i][j] += val; } __int64 sum(int row, int col) { __int64 res = 0; int i, j; for(i = row; i >= 1; i -= lowbit(i)) for(j = col; j >= 1; j -= lowbit(j)) res += countv[i][j]; return res; } int main() { int i, j, pos1, pos2; scanf("%d", &caseNum); for(i = 1; i <= caseNum; i++) { total = 0; //memset(countv, 0, sizeof(countv)); scanf("%d%d%d", &numN, &numM, &numK); for(pos1 = 1; pos1 <= numN; pos1++) for(pos2 = 1; pos2 <= numM; pos2++) countv[pos1][pos2] = 0; for(j = 1; j <= numK; j++) { scanf("%d%d", &pos1, &pos2); __int64 res1 = sum(pos1 - 1, numM); __int64 res2 = sum(pos1 - 1, pos2); __int64 res3 = sum(numN, pos2 - 1); __int64 res4 = sum(pos1, pos2 - 1); total += res1 - res2 + res3 - res4; modify(pos1, pos2, 1); } printf("Test case %d: %I64d/n", i, total); } return 0; } */ //一维树状数组 #include <iostream> #include <algorithm> #define MAX_N 1000 using namespace std; __int64 countv[MAX_N + 1]; int numN, numM, numK, caseNum; __int64 total; struct input { int rowId, colId; }inputs[MAX_N * MAX_N + 5]; bool compare(const input p1, const input p2) { if(p1.rowId < p2.rowId) return true; else if(p1.rowId == p2.rowId) return p1.colId <= p2.colId; else return false; } int lowbit(int p) { return p & (p ^ (p - 1)); } void modify(int pos, __int64 val) { int i; for(i = pos; i <= numM; i += lowbit(i)) countv[i] += val; } __int64 sum(int pos) { __int64 res = 0; int i; for(i = pos; i >= 1; i -= lowbit(i)) res += countv[i]; return res; } int main() { int i, j, pos1, pos2; scanf("%d", &caseNum); for(i = 1; i <= caseNum; i++) { total = 0; //memset(countv, 0, sizeof(countv)); scanf("%d%d%d", &numN, &numM, &numK); for(pos2 = 1; pos2 <= numM; pos2++) countv[pos2] = 0; for(j = 0; j < numK; j++) { scanf("%d%d", &pos1, &pos2); inputs[j].rowId = pos1; inputs[j].colId = pos2; } sort(inputs, inputs + numK, compare); for(j = 0; j < numK; j++) { pos2 = inputs[j].colId; __int64 res1 = sum(numM); __int64 res2 = sum(pos2); total += res1 - res2; modify(pos2, 1); } printf("Test case %d: %I64d/n", i, total); } return 0; }