BZOJ 3389: [Usaco2004 Dec]Cleaning Shifts安排值班

题目

3389: [Usaco2004 Dec]Cleaning Shifts安排值班

Time Limit: 1 Sec Memory Limit: 128 MB

Description

一天有T(1≤T≤10^6)个时段．约翰正打算安排他的N(1≤N≤25000)只奶牛来值班，打扫

打扫牛棚卫生．每只奶牛都有自己的空闲时间段[Si，Ei](1≤Si≤Ei≤T)，只能把空闲的奶牛安排出来值班．而且，每个时间段必需有奶牛在值班．那么，最少需要动用多少奶牛参与值班呢？如果没有办法安排出合理的方案，就输出-1.

Input

第1行：N，T．

第2到N+1行：Si，Ei．

Output

最少安排的奶牛数．

Sample Input

3 10
1 7
3 6
6 10

Sample Output

2

样例说明
奶牛1和奶牛3参与值班即可．

HINT

Source

Silver

题解

呵呵，做了一半忽然发现这不是昨天做的那题（BZOJ 1672）的减弱版，题解直接见http://www.cnblogs.com/WNJXYK/p/4074788.html。线段树动态规划！【听说贪心可做，Orz但那是我N^2贪心TLE了两次、、、

代码

  1 /*Author:WNJXYK*/
  2 #include<cstdio>
  3 #include<algorithm>
  4 using namespace std;
  5 
  6 int n,st,ed;
  7 struct line{
  8     int left,right;
  9     int w;
 10 }cow[25010];
 11 bool cmp(line a,line b){
 12     if (a.left<b.left) return true;
 13     if (a.left==b.left && a.right<b.right) return true;
 14     return false;
 15 }
 16 inline int remin(int a,int b){
 17     if (a<b) return a;
 18     return b;    
 19 }
 20 inline int remax(int a,int b){
 21     if (a>b) return a;
 22     return b;
 23 }
 24 
 25 const int Maxn=1000000;
 26 const int Inf=2000000000;
 27 struct Btree{
 28     int left,right;
 29     int min;
 30     int tag;
 31 }tree[Maxn*4+10];
 32 
 33 void build(int x,int left,int right){
 34     tree[x].left=left;
 35     tree[x].right=right;
 36     tree[x].tag=Inf;
 37     if (left==right){
 38         tree[x].min=(left<st?0:Inf);
 39     }else{
 40         int mid=(left+right)/2;
 41         build(x*2,left,mid);
 42         build(x*2+1,mid+1,right);
 43         tree[x].min=remin(tree[x*2].min,tree[x*2+1].min);
 44     }
 45 }
 46 
 47 inline void clean(int x){
 48     if (tree[x].left!=tree[x].right){
 49         tree[x*2].min=remin(tree[x].tag,tree[x*2].min);
 50         tree[x*2].tag=remin(tree[x].tag,tree[x*2].tag);
 51         tree[x*2+1].min=remin(tree[x].tag,tree[x*2+1].min);
 52         tree[x*2+1].tag=remin(tree[x].tag,tree[x*2+1].tag);
 53         tree[x].tag=Inf;
 54     }
 55 }
 56 
 57 void change(int x,int left,int right,int val){
 58     clean(x);
 59     if (left<=tree[x].left && tree[x].right<=right){
 60         tree[x].tag=remin(tree[x].tag,val);
 61         tree[x].min=remin(tree[x].min,val);
 62     }else{
 63         int mid=(tree[x].left+tree[x].right)/2;
 64         if (left<=mid) change(x*2,left,right,val);
 65         if (right>=mid+1)change(x*2+1,left,right,val);
 66         tree[x].min=remin(tree[x*2].min,tree[x*2+1].min);
 67     }
 68 }
 69 
 70 int query(int x,int left,int right){
 71     clean(x);
 72     if (left<=tree[x].left && tree[x].right<=right){
 73         return tree[x].min;
 74     }else{
 75         int Ans=Inf;
 76         int mid=(tree[x].left+tree[x].right)/2;
 77         if (left<=mid) Ans=remin(Ans,query(x*2,left,right));
 78         if (right>=mid+1) Ans=remin(Ans,query(x*2+1,left,right));
 79         return Ans; 
 80     }
 81 }
 82 
 83 
 84 int main(){
 85     scanf("%d%d",&n,&ed);
 86     st=1;
 87     build(1,0,ed);
 88     for (int i=1;i<=n;i++){
 89         scanf("%d%d",&cow[i].left,&cow[i].right);
 90         cow[i].w=1;
 91     }
 92     sort(cow+1,cow+n+1,cmp);
 93     for (int i=1;i<=n;i++){
 94         int mindist=query(1,remax(cow[i].left-1,0),cow[i].right)+cow[i].w;
 95         //printf("mindist:%d
",mindist);
 96         //printf("query %d %d -> min=%d
",remax(cow[i].left-1,0),cow[i].right,query(1,cow[i].left,cow[i].right));
 97         change(1,cow[i].left,cow[i].right,mindist);
 98     } 
 99         //printf("query min=%d
",query(1,ed,ed));
100     int ans=query(1,ed,ed);
101     if (ans==Inf)
102         printf("-1
");
103     else
104         printf("%d
",ans); 
105     return 0;
106 }

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