【POJ 1716 Integer Intervals】+ 差分约束

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

4

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;
}