HDU 6301 贪心

Distinct Values

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2321    Accepted Submission(s): 748

Problem Description

Chiaki has an array of n positive integers. You are told some facts about the array: for every two elements ai and aj in the subarray al..r (l≤i<j≤r), ai≠aj holds.
Chiaki would like to find a lexicographically minimal array which meets the facts.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers n and m (1≤n,m≤105) -- the length of the array and the number of facts. Each of the next m lines contains two integers li and ri (1≤li≤ri≤n).

It is guaranteed that neither the sum of all n nor the sum of all m exceeds 106.

 

Output

For each test case, output n integers denoting the lexicographically minimal array. Integers should be separated by a single space, and no extra spaces are allowed at the end of lines.

 

Sample Input

3

2 1

1 2

4 2

1 2

3 4

5 2

1 3

2 4

 

Sample Output

1 2

1 2 1 2

1 2 3 1 1

 

Source

2018 Multi-University Training Contest 1

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=6301

解析 先把区间排序 左端点靠左 右端点靠右 然后就可以根据右端点是否递增进行 贪心 然后就可以了。

memset也比较耗时 所以就用多少开多少。

AC代码

#include<stdio.h>
#include<string.h>
#include<set>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e6+20,mod=1e9+7,inf=0x3f3f3f3f;
typedef long long ll;
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("
");
#define debug(a,b) cout<<a<<" "<<b<<" ";
/*================================================================================*/
struct node
{
    int l,r;
    bool operator <(const node &a)const
    {
        if(l==a.l)
            return r>a.r;
        return l<a.l;
    }
}a[maxn];
//int ans[maxn];
int main()
{
    int n,m,t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%d%d",&n,&m);
            vector<int> ans(n+1,1);
           //fillchar(ans,0);
            for(int i=0;i<m;i++)
                scanf("%d%d",&a[i].l,&a[i].r);
            sort(a,a+m);
            set<int> s;
            for(int i=1;i<=n;i++) s.insert(i);
            int l=1,r=0,flag=0;
            for(int i=0;i<m;i++)
            {
                if(a[i].r<r)           //右端点小于r就不用管了 
                    continue;
                else
                {
                    if(i!=0)
                    {

                        for(int k=a[flag].l;k<=min(a[i].l-1,r);k++) //将之前用过的数 且不在当前区间的数加入set
                            s.insert(ans[k]);
                        l=max(r+1,a[i].l);
                    }
                    else                        //更新l r
                        l=a[i].l;
                    r=a[i].r;                                      
                    flag=i;   //上一个进行贪心的区间
                }
                for(int j=l;j<=r;j++)
                {
                    ans[j]=*s.begin();s.erase(ans[j]); //把用过的从set删除
                }
            }
            for(int i=1;i<=n;i++)
            {
                //if(!ans[i])ans[i]=1;
                if(i!=1)
                    printf(" %d",ans[i]);
                else
                    printf("%d",ans[i]);
            }
            printf("
");
        }
    }
}