Poj 3067 Japan (line segment tree), poj3067

Poj 3067 Japan (line segment tree), poj3067

Link: poj 3067 Japan

Given N and M, it indicates the number of cities in the East and West, and then K railways, each railway connects east and west cities, and asks how many intersections there will be.

Solution: The line segment tree can be maintained. Each side is sorted by x smaller and y smaller, and then y + 1 to M can be queried each time.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1005;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)|1)int lc[maxn << 2], rc[maxn << 2], s[maxn << 2];inline void pushup(int u) {    s[u] = s[lson(u)] + s[rson(u)];}void build (int u, int l, int r) {    lc[u] = l;    rc[u] = r;    s[u] = 0;    if (l == r)        return ;    int mid = (l + r) >> 1;    build(lson(u), l, mid);    build(rson(u), mid + 1, r);    pushup(u);}void modify(int u, int x, int w) {    if (x == lc[u] && rc[u] == x) {        s[u] += w;        return;    }        int mid = (lc[u] + rc[u]) >> 1;    if (x <= mid)        modify(lson(u), x, w);    else        modify(rson(u), x, w);    pushup(u);}int query(int u, int l, int r) {    if (l > r)        return 0;    if (l <= lc[u] && rc[u] <= r)        return s[u];    int mid = (lc[u] + rc[u]) >> 1, ret = 0;    if (l <= mid)        ret += query(lson(u), l, r);    if (r > mid)        ret += query(rson(u), l, r);    return ret;}int N, M, Q;struct point {    int x, y;    friend bool operator < (const point& a, const point& b) {        if (a.x != b.x)            return a.x < b.x;        return a.y < b.y;    }}p[maxn * maxn];int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        scanf("%d%d%d", &N, &M, &Q);        build(1, 1, M);        long long ans = 0;        for (int i = 0; i < Q; i++)            scanf("%d%d", &p[i].x, &p[i].y);        sort(p, p + Q);        for (int i = 0; i < Q; i++) {            ans += query(1, p[i].y + 1, M);            modify(1, p[i].y, 1);        }        printf("Test case %d: %lld\n", kcas, ans);    }    return 0;}