hdu 6301 Distinct Values（贪心+优先队列）

最新推荐文章于 2020-01-11 20:46:59 发布

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本文介绍了一道算法题目，要求在给定序列长度及区间约束条件下，找到满足区间内元素互不相同的字典序最小的序列。通过贪心算法和优先队列实现了区间内未出现过的最小正整数的选择。

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【题目】

Distinct Values

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4054 Accepted Submission(s): 1360

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

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

【题解】

题意：给定序列长度n和m个区间，要求给定的区间内要素不同，求字典序最小的序列。

思路：运用贪心的思想把问题转化为寻找区间内最小的没有出现过的正整数。排序处理下区间，用优先队列维护即可。

【代码】

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
struct p{
    int l,r;
}f[maxn];
int ans[maxn];
bool cmp(p a,p b)
{
    if(a.l==b.l)
        return a.r>b.r;
    return a.l<b.l;
}
int main()
{
    int t; scanf("%d",&t);
    while(t--)
    {
        priority_queue <int,vector<int>,greater<int> > q;
        int n,m; scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            q.push(i),ans[i]=1;
        for(int i=0;i<m;i++)
            scanf("%d%d",&f[i].l,&f[i].r);
        sort(f,f+m,cmp);
        int bg=1,ed=1;
        q.pop();
        for(int i=0;i<m;i++){
            if(f[i].l>=bg&&f[i].r<=ed) continue;
            if(f[i].l>ed){
                for(int j=bg;j<=ed;j++)
                    q.push(ans[j]);
                for(int j=f[i].l;j<=f[i].r;j++)
                    ans[j]=q.top(),q.pop();
            }
            else{
                for(int j=bg;j<f[i].l;j++)
                    q.push(ans[j]);
                for(int j=ed+1;j<=f[i].r;j++)
                    ans[j]=q.top(),q.pop();
            }
            bg=f[i].l;
            ed=f[i].r;
        }
        for(int i=1;i<=n;i++)
            printf(i<n?"%d ":"%d\n",ans[i]);
    }
    return 0;
}