Ultraviolet A 1511 soju (Greedy)

Ultraviolet A 1511 Soju

Question Link

Given two point sets, the two points set must take the minimum Manhattan distance from each other to ensure that X of Point Set 1 is less than 0 and X of Point Set 2 is greater than 0.

Idea: Because X2> X1, you only need to consider the Y value. If one y is larger than the other y, It is Y1-Y2. Otherwise, it is Y2-Y1, in this way, you only need to perform two sort greedy computations for these two cases.

Code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int N = 200005;int t, n, m, a1, a2, a3, a4;struct Point {    int x, y, flag;} p[N];bool cmp(Point a, Point b) {    return a.y < b.y;}int solve() {    int ans = INF;    sort(p, p + n + m, cmp);    int tmp = INF;    for (int i = 0; i < n + m; i++) {if (p[i].flag)    tmp = min(tmp, p[i].x - p[i].y);else    ans = min(ans, p[i].y - p[i].x + tmp);    }    tmp = INF;    reverse(p, p + n + m);    for (int i = 0; i < n + m; i++) {if (p[i].flag)    tmp = min(tmp, p[i].x + p[i].y);else    ans = min(ans, -p[i].x - p[i].y + tmp);    }    return ans;}int main() {    scanf("%d", &t);    while (t--) {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 = n; i < n + m; i++) {    scanf("%d%d", &p[i].x, &p[i].y);    p[i].flag = 1;}printf("%d\n", solve());    }    return 0;}