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 MBDescription
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
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\n",mindist); 96 //printf("query %d %d -> min=%d\n",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\n",query(1,ed,ed)); 100 int ans=query(1,ed,ed); 101 if (ans==Inf) 102 printf("-1\n"); 103 else 104 printf("%d\n",ans); 105 return 0; 106 }
