题目描述

You are given an integer sequence x of length N. Determine if there exists an integer sequence a that satisfies all of the following conditions, and if it exists, construct an instance of a.

a is N2 in length, containing N copies of each of the integers 1, 2, …, N.
For each 1≤i≤N, the i-th occurrence of the integer i from the left in a is the xi-th element of a from the left.
Constraints
1≤N≤500
1≤xi≤N 2
All xi are distinct.

输入

The input is given from Standard Input in the following format:

N
x1 x2 … xN

输出

If there does not exist an integer sequence a that satisfies all the conditions, print 'No'. If there does exist such an sequence a, print 'Yes'.

样例输入

3
1 5 9

样例输出

Yes
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int a[255000];
struct node
{
    int num;
    int x;
}s[2550];
bool cmp(node a,node b)
{
    return a.x<b.x;
}
int main()
{
    int i,j,n;
    cin>>n;
    for(i=1;i<=n;i++){
        cin>>s[i].x;
        s[i].num=i;
    }
    sort(s+1,s+n+1,cmp);
    int now=1;
    for(i=1;i<=n;i++){
        a[s[i].x]=s[i].num;
        for(j=1;j<s[i].num;j++){
            while(a[now]) now++;
            a[now]=s[i].num;
        }
        if(now>s[i].x) {cout<<"No"<<endl;return 0;}
    }
    for(i=1;i<=n;i++){
        for(j=1;j<=n-s[i].num;j++){
            while(a[now]) now++;
            if(now<s[i].x) {cout<<"No"<<endl;return 0;}
            a[now]=s[i].num;
        }
    }
    cout<<"Yes"<<endl;
    return 0;
}
View Code

For each 1≤i≤N, the i-th occurrence of the integer i from the left in a is the xi-th element of a from the left.

就这句话。。。仔细读

第i个i在下标为xi的地方。嗯

这是贪心，先放上这些数字应在的位置

再从小到大把i弄上去

排除不可能的解