題目鏈接

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

題意

日本有東西兩個海岸，東海岸有N個城市，西海岸有M個城市。城市之間一共修了K條路，問一共有幾個交點。
當兩條路A、B，A的左端點小于B的左端點，A的右端點大于B的右端點時相交。
所以先按照左端點排序，剩下的就是求逆序數。

AC

• 樹狀數組（記得開long long）

#include <iostream> #include <algorithm> #include <stdio.h> #include <vector> #include <map> #include <bitset> #include <set> #include <string.h> #include <cmath> #include <queue> #include <algorithm> #define N 1005 #define P pair<int,int> #define ll long long #define lowbit(a) a&(-a) #define pb(a) push_back(a) #define mk(a, b) make_pair(a, b) #define mem(a, b) memset(a, b, sizeof(a)) #define read(a) scanf("%d", &a) #define print(a) printf("%d\n", a) using namespace std; ll c[N]; ll getsum(int x) {ll ret = 0;while (x) {ret += c[x];x -= lowbit(x);}return ret; } void update(int x) {while (x < N) {c[x]++;x += lowbit(x);} } int main(){ #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endifint t;scanf("%d", &t);int Case = 1;while (t--) {mem(c, 0);int n, m, k;scanf("%d%d%d", &n, &m, &k);vector<P> a(k);for (int i = 0; i < k; ++i) {scanf("%d%d", &a[i].first, &a[i].second);}// 按照左端點升序，左端點相同又端點升序sort(a.begin(), a.end());ll ans = 0;for (int i = 0; i < k; ++i) {ans += getsum(N) - getsum(a[i].second);update(a[i].second);}printf("Test case %d: %lld\n", Case++, ans);}return 0; }

• 線段樹

#include <iostream> #include <algorithm> #include <stdio.h> #include <vector> #include <map> #include <bitset> #include <set> #include <string.h> #include <cmath> #include <queue> #include <algorithm> #define N 4005 #define P pair<int,int> #define ll long long #define lowbit(a) a&(-a) #define mk(a, b) make_pair(a, b) #define mem(a, b) memset(a, b, sizeof(a)) using namespace std;struct node{ll val; }segtree[N]; void build(int root, int start, int end) {if (start == end)segtree[root].val = 0;else {int mid = (start + end) / 2;build(root * 2 + 1, start, mid);build(root * 2 + 2, mid + 1, end);segtree[root].val = segtree[root * 2 + 1].val + segtree[root * 2 + 2].val;} }void update(int root, int start, int end, int index) {if (index > end || index < start)return;if (start == end) {if (start == index) {segtree[root].val++;}return;}int mid = (start + end) / 2;update(root * 2 + 1, start, mid, index);update(root * 2 + 2, mid + 1, end, index);segtree[root].val = segtree[root * 2 + 1].val + segtree[root * 2 + 2].val; } ll query(int root, int start, int end, int qstart, int qend) {if (qstart > end || qend < start)return 0;if (start >= qstart && end <= qend) {return segtree[root].val;}int mid = (start + end) / 2;return query(root * 2 + 1, start, mid, qstart, qend) + query(root * 2 + 2, mid + 1, end, qstart, qend); }int main(){ #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endifint t;scanf("%d", &t);int Case = 1;while (t--) {int n, m, k;scanf("%d%d%d", &n, &m, &k);vector<P> a(k);for (int i = 0; i < k; ++i) {scanf("%d%d", &a[i].first, &a[i].second);}// 按照左端點升序，左端點相同又端點升序sort(a.begin(), a.end());build(0, 0, m + 1);ll ans = 0;for (int i = 0; i < k; ++i) {// 如果查詢當前點，可能會被前面的區間包含ans += query(0, 0, m + 1, a[i].second + 1, m + 1);update(0, 0, m + 1, a[i].second);}printf("Test case %d: %lld\n", Case++, ans);}return 0; }

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