Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 32561 | Accepted: 7972 |
Description
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
* 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
Sample Input
3 10
1 7
3 6
6 10
Sample Output
2
Hint
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.
Source
https://www.cnblogs.com/wyboooo/p/9808378.html
和poj3171基本一样,改一下输入和范围即可。
#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f const int maxn = + ;
const int maxtime = 1e6 + ;
struct node{
int st, ed, cost;
}cow[maxn];
bool cmp(node a, node b)
{
return a.ed < b.ed;
}
LL tree[maxtime << ];//区间中f[]最小值
int n, L, R; void pushup(int rt)
{
tree[rt] = min(tree[rt << ], tree[rt << |]);
} void build(int rt, int l, int r)
{
if(l == r){
tree[maxn] = inf;
return;
}
int mid = (l + r) / ;
build(rt<<, l, mid);
build(rt<<|, mid + , r);
pushup(rt);
} void update(int x, LL val, int l, int r, int rt)
{
if(l == r){
tree[rt] = min(tree[rt], val);
return;
}
int m = (l + r) / ;
if(x <= m){
update(x, val, l, m, rt<<);
}
else{
update(x, val, m + , r, rt<<|);
}
pushup(rt);
} LL query(int L, int R, int l, int r, int rt)
{
if(L <= l && R >= r){
return tree[rt];
}
int m = (l + r) / ;
LL ans = inf;
if(L <= m){
ans = min(ans, query(L, R, l, m, rt<< ));
}
if(R > m){
ans = min(ans, query(L, R, m + , r, rt<<|));
}
pushup(rt);
return ans;
} int main()
{
while(scanf("%d%d", &n, &R) != EOF){
R+=;
memset(tree, 0x7f, sizeof(tree));
for(int i = ; i <= n; i++){
scanf("%d%d", &cow[i].st, &cow[i].ed);
cow[i].st+=;cow[i].ed+=;
cow[i].cost = ;
}
sort(cow + , cow + + n, cmp); build(, , R); update(, , , R, );
//cout<<"yes"<<endl;
//int far = L;
bool flag = true;
for(int i = ; i <= n; i++){
/*if(cow[i].st > far + 1){
flag = false;
// break;
}*/
int a = max(, cow[i].st - );
int b = min(R, cow[i].ed);
//cout<<a<<" "<<b<<endl;
LL f = query(a, b, , R, );
f += cow[i].cost;
//cout<<f<<endl;
update(b, f, , R, );
//far = max(far, cow[i].ed);
//cout<<far<<endl;
}
//cout<<"yes"<<endl; LL ans = query(R, R, , R, );
if(ans >= inf){
printf("-1\n");
}
else{
printf("%lld\n", ans); //else{
// printf("-1\n");
} } }