题目链接：uva 1511 - Soju

题目大意：给出两个点集，问说分别从两个点集中取一点的哈夫曼距离最小值。注意一个点集的x坐标小于0，另一个大于0.

解题思路：因为x2一定大于x1，所以对于x这一维，一定是+x2-x1，所以只需要考虑y这一维坐标即可。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;
const int maxn = 1e5+5;
const int INF = 0x3f3f3f3f;

struct point {
    int x, y, flag;
}p[maxn*2];
int N, M;

bool cmp (const point& a, const point& b) {
    return a.y < b.y;
}

void init () {
    scanf("%d", &N);
    for (int i = 0; i < N; i++) {
        scanf("%d%d", &p[i].x, &p[i].y);
        p[i].flag = 0;
    }

    scanf("%d", &M);
    for (int i = 0; i < M; i++) {
        scanf("%d%d", &p[i+N].x, &p[i+N].y);
        p[i+N].flag = 1;
    }

}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();
        int ans = INF, now = INF;
        sort(p, p + N + M, cmp);
        for (int i = 0; i < N + M; i++) {
            if (p[i].flag)
                now = min(now, p[i].x - p[i].y);
            else
                ans = min(ans, now - p[i].x + p[i].y);
        }

        now = INF;
        reverse(p, p + N + M);
        for (int i = 0; i < N + M; i++) {
            if (p[i].flag)
                now = min(now, p[i].x + p[i].y);
            else
                ans = min(ans, now - p[i].x - p[i].y);
        }
        printf("%d\n", ans);
    }
    return 0;
}