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

 /*Author:WNJXYK*/
 #include<cstdio>
 #include<algorithm>
 using namespace std;

 int n,st,ed;
 struct line{
     int left,right;
     int w;
 }cow[];
 bool cmp(line a,line b){
     if (a.left<b.left) return true;
     if (a.left==b.left && a.right<b.right) return true;
     return false;
 }
 inline int remin(int a,int b){
     if (a<b) return a;
     return b;
 }
 inline int remax(int a,int b){
     if (a>b) return a;
     return b;
 }

 const int Maxn=;
 const int Inf=;
 struct Btree{
     int left,right;
     int min;
     int tag;
 }tree[Maxn*+];

 void build(int x,int left,int right){
     tree[x].left=left;
     tree[x].right=right;
     tree[x].tag=Inf;
     if (left==right){
         tree[x].min=(left<st?:Inf);
     }else{
         int mid=(left+right)/;
         build(x*,left,mid);
         build(x*+,mid+,right);
         tree[x].min=remin(tree[x*].min,tree[x*+].min);
     }
 }

 inline void clean(int x){
     if (tree[x].left!=tree[x].right){
         tree[x*].min=remin(tree[x].tag,tree[x*].min);
         tree[x*].tag=remin(tree[x].tag,tree[x*].tag);
         tree[x*+].min=remin(tree[x].tag,tree[x*+].min);
         tree[x*+].tag=remin(tree[x].tag,tree[x*+].tag);
         tree[x].tag=Inf;
     }
 }

 void change(int x,int left,int right,int val){
     clean(x);
     if (left<=tree[x].left && tree[x].right<=right){
         tree[x].tag=remin(tree[x].tag,val);
         tree[x].min=remin(tree[x].min,val);
     }else{
         int mid=(tree[x].left+tree[x].right)/;
         if (left<=mid) change(x*,left,right,val);
         if (right>=mid+)change(x*+,left,right,val);
         tree[x].min=remin(tree[x*].min,tree[x*+].min);
     }
 }

 int query(int x,int left,int right){
     clean(x);
     if (left<=tree[x].left && tree[x].right<=right){
         return tree[x].min;
     }else{
         int Ans=Inf;
         int mid=(tree[x].left+tree[x].right)/;
         if (left<=mid) Ans=remin(Ans,query(x*,left,right));
         if (right>=mid+) Ans=remin(Ans,query(x*+,left,right));
         return Ans;
     }
 }

 int main(){
     scanf("%d%d",&n,&ed);
     st=;
     build(,,ed);
     for (int i=;i<=n;i++){
         scanf("%d%d",&cow[i].left,&cow[i].right);
         cow[i].w=;
     }
     sort(cow+,cow+n+,cmp);
     for (int i=;i<=n;i++){
         int mindist=query(,remax(cow[i].left-,),cow[i].right)+cow[i].w;
         //printf("mindist:%d\n",mindist);
         //printf("query %d %d -> min=%d\n",remax(cow[i].left-1,0),cow[i].right,query(1,cow[i].left,cow[i].right));
         change(,cow[i].left,cow[i].right,mindist);
     }
         //printf("query min=%d\n",query(1,ed,ed));
     int ans=query(,ed,ed);
     if (ans==Inf)
         printf("-1\n");
     else
         printf("%d\n",ans);
     return ;
 }