poj 3067 Japan(线段树)

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题目链接：poj 3067 Japan

题目大意：给定N和M，表示东部和西部城市的数量，然后K条铁路，每条铁路连接东西城市，问说会有多少次交点。

解题思路：线段树维护即可，每条边按照x小的，y小的排序，然后每次查询y+1到M的即可。

#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;
}

poj 3067 Japan(线段树)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/40409443