Cleaning Shifts
Description
Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.
Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.
Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1
Input
* Line 1: Two space-separated integers: N and T
* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.
*
Output
* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.
*
Sample Input
3 10
1 7
3 6
6 10
Sample Output
2
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
INPUT DETAILS:
There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10.
OUTPUT DETAILS:
By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.
题意：一个区间长度为T，给出N个区间，求至少需要多少个区间能完全覆盖此区间。
分析：dp方程很容易想到，在此不赘述，设f[i]表示覆盖[L,i]最少需要的区间数，f[i]=min(f[x])+1,a[i]-1<=x<=b[i]。直接枚举显然超时，需要用线段树维护区间[a[i]-1,b[i]]的最小值，方便查询。
代码
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#define N 1000005
#define inf 1e9
using namespace std;
struct tree
{
int l, r, min;
}tr[N * 4],a[N];
int n,m,f[N],x;
int cmp(tree x, tree y){return x.r < y.r;}
void build(int p)
{
tr[p].min = inf;
if (tr[p].l == tr[p].r)
{
if (tr[p].l == 0) tr[p].min = 0;
return;
}
int mid = (tr[p].l + tr[p].r) / 2;
tr[p * 2].l = tr[p].l;
tr[p * 2].r = mid;
tr[p * 2 + 1].l = mid + 1;
tr[p * 2 + 1].r = tr[p].r;
build(p * 2);
build(p * 2 + 1);
tr[p].min = min(tr[p * 2].min, tr[p * 2 + 1].min);
}
void find(int p, int L, int R)
{
if (tr[p].l == L && tr[p].r == R)
{
x = min(x, tr[p].min);
return;
}
int mid = (tr[p].l + tr[p].r) / 2;
if (R <= mid) find(p * 2, L, R);
else if (L > mid) find(p * 2 + 1, L, R);
else find(p * 2, L, mid), find(p * 2 + 1, mid + 1, R);
}
void ins(int p, int pos, int val)
{
if (tr[p].l == tr[p].r)
{
tr[p].min = min(tr[p].min, val);
return;
}
int mid = (tr[p].l + tr[p].r) / 2;
if (pos <= mid) ins(p * 2, pos, val);
else ins(p * 2 + 1, pos, val);
tr[p].min = min(tr[p * 2].min, tr[p * 2 + 1].min);
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d%d", &a[i].l, &a[i].r);
sort(a + 1, a + n + 1, cmp);
tr[1].l = 0;
tr[1].r = m;
build(1);
for (int i = 1; i <= m; i++)
f[i] = inf;
for (int i = 1; i <= n; i++)
{
x = inf;
if (a[i].l < 1) a[i].l = 1;
if (a[i].r > m) a[i].r = n;
find(1, a[i].l - 1, a[i].r);
if (x + 1 < f[a[i].r])
{
f[a[i].r] = x + 1;
ins(1, a[i].r, f[a[i].r]);
}
}
if (f[m] < inf) printf("%d", f[m]);
else printf("-1");
}