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来源：互联网发布：软件项目范围管理编辑：程序博客网时间：2024/05/16 14:39

题意:

点分为两个区,一个区的横坐标小于第二个区,求abs(x1-x2)+abs(y1-y2)的最小值

思路:

主要思想就是把下标一样的同一到一边,那就要分类讨论

由于x1<x2所以就是x2-x1,然后分类讨论y1,y2的情况,所以循环两次,当然两边是需要同时维护的

代码:

#include<cstdio>#include<string>#include<algorithm>using namespace std;const int maxn = 100005;const int INF = 0x3f3f3f3f;int n, m;struct point {int x, y;int flag;}p[maxn*2];int cmp(point a, point b) {return a.y < b.y;}void deal() {int temp = INF, ans = INF;//分两次讨论for(int i=0; i<n+m; i++) {if(p[i].flag) {//y2<y1temp = min(temp, p[i].x-p[i].y);}else ans = min(ans, temp+p[i].y-p[i].x);}temp = INF;for(int i=0; i<n+m; i++) {if(!p[i].flag) {temp = min(temp, -(p[i].x+p[i].y));}else {ans = min(ans, temp+p[i].x+p[i].y);}}printf("%d\n", ans);}int main() {int kase;scanf("%d", &kase);while(kase--) {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;}sort(p, p+n+m, cmp);deal();}return 0;}

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