【POJ 1716 Integer Intervals】+ 差分约束

最新推荐文章于 2020-06-17 12:30:24 发布

楚江枫

最新推荐文章于 2020-06-17 12:30:24 发布

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本文介绍了一种算法，用于找出包含每个指定区间至少两个不同整数的最小元素集合。输入为一系列整数区间，输出是最小集合的元素数量。

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Integer Intervals
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 14475 Accepted: 6149
Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b.
Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.
Input

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an interval.
Output

Output the minimal number of elements in a set containing at least two different integers from each interval.
Sample Input

4
3 6
2 4
0 2
4 7
Sample Output

限制条件有三个：

1、对于区间，必定有num[start] <= num[end]-2;

2、对于整数点，必有num[i] <= num[i+1];

3、对于整数点，必有num[i+1] <= num[i]+1;

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;
const int K = 1e4 + 10;
int a[K],b[K],num[K];
int main()
{
    int N;
    while(scanf("%d",&N) != EOF){
        int l = K,r = 0;
        for(int i = 1; i <= N; i++)
            scanf("%d %d",&a[i],&b[i]), b[i]++,l = min(l,a[i]),r = max(r,b[i]);
        int ok = 1;
        while(ok){
          ok = 0;
          for(int i = 1; i <= N; i++)
             if(num[a[i]] > num[b[i]] - 2)
                num[a[i]] = num[b[i]] - 2,ok = 1;
          for(int i = l; i < r; i++)
             if(num[i + 1] > num[i] + 1)
                num[i + 1] = num[i] + 1,ok = 1;
          for(int i = r - 1; i >= l; i--)
            if(num[i] > num[i + 1])
                num[i] = num[i + 1],ok = 1;
         }
         printf("%d\n",num[r] - num[l]);
    }
    return 0;
}