Hdu 3552 I can do it!(贪心）

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I can do it!

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1060 Accepted Submission(s): 498

Problem Description

Given n elements, which have two properties, say Property A and Property B. For convenience, we use two integers A_i and B_i to measure the two properties.
Your task is, to partition the element into two sets, say Set A and Set B , which minimizes the value of max_{(x∈Set A)} {A_x}+max_{(y∈Set B)} {B_y}.
See sample test cases for further details.

Input

There are multiple test cases, the first line of input contains an integer denoting the number of test cases.
For each test case, the first line contains an integer N, indicates the number of elements. (1 <= N <= 100000)
For the next N lines, every line contains two integers A_i and B_i indicate the Property A and Property B of the ith element. (0 <= A_i, B_i <= 1000000000)

Output

For each test cases, output the minimum value.

Sample Input

131 1002 1003 1

Sample Output

Case 1: 3
 
做这道题的时候，思维很乱很乱，所以捋了好久，题意就是有n个元素，每个元素有x，y两个属性，让你把这些元素分成两个集合A和B，让A中x属性的最大值和B中y属性的最大值加和最小。可以想到，将n个元素按照x属性的值排序后，如果A集合中x的属性最大值确定的话，那么和它相加的那个值一定是剩下元素中y属性的最大值，因为说白了x属性的最大值就是A和B集合的分界线，A都分出来了那剩下的就是B，所以我们枚举A的x属性最大值就行了
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int maxn = 100005;const int Inf = 0x3f3f3f;int t,n;struct Node{    int x;    int y;}node[maxn];int Sort[maxn];int cmp(Node a,Node b){    return a.x < b.x;}int main(){    scanf("%d",&t);    int cnt = 0;    while(t--)    {        cnt++;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d%d",&node[i].x,&node[i].y);        }        sort(node,node+n,cmp);        int max = -1;        int min = Inf;        for(int i=n-1;i>=0;i--)        {            if(max < node[i].y)                max = node[i].y;            Sort[i] = max;        }        for(int i=0;i<n-1;i++)        {            int sum = node[i].x + Sort[i+1];            if(min > sum)                min = sum;        }        printf("Case %d: ",cnt);        printf("%d\n",min);    }}

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