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HDU4614 Vases and Flowers

时间:2016-07-03 21:21:22      阅读:209      评论:0      收藏:0      [点我收藏+]

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http://acm.hdu.edu.cn/showproblem.php?pid=4614

 HDU 4614 Vases and Flowers (2013多校第二场线段树)

  1 // #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include <iostream>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <sstream>
  6 #include <string>
  7 #include <algorithm>
  8 #include <list>
  9 #include <map>
 10 #include <vector>
 11 #include <queue>
 12 #include <stack>
 13 #include <cmath>
 14 #include <cstdlib>
 15 #include <conio.h>
 16 using namespace std;
 17 #define clc(a,b) memset(a,b,sizeof(a))
 18 #define inf 0x3f3f3f3f
 19 #define lson l,mid,rt<<1
 20 #define rson mid+1,r,rt<<1|1
 21 const int N = 50010;
 22 const int MOD = 1e9+7;
 23 #define LL long long
 24 #define mi() (l+r)>>1
 25 double const pi = acos(-1);
 26 void fre() {
 27     freopen("in.txt","r",stdin);
 28 }
 29 // inline int r() {
 30 //     int x=0,f=1;char ch=getchar();
 31 //     while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1;ch=getchar();}
 32 //     while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘;ch=getchar();}return x*f;
 33 // }
 34 struct Edge {
 35     int l,r;
 36     int lazy,sum;
 37 } e[N<<2];
 38 
 39 void pushdown(int rt) {
 40     e[rt].sum=e[rt<<1].sum+e[rt<<1|1].sum;
 41 }
 42 void pushup(int rt) {
 43     if(e[rt].lazy!=-1) {
 44         e[rt<<1].lazy=e[rt<<1|1].lazy=e[rt].lazy;
 45         e[rt<<1].sum=(e[rt<<1].r-e[rt<<1].l+1)*e[rt].lazy;
 46         e[rt<<1|1].sum=(e[rt<<1|1].r-e[rt<<1|1].l+1)*e[rt].lazy;
 47         e[rt].lazy=-1;
 48     }
 49 }
 50 
 51 void build(int l,int r,int rt) {
 52     e[rt].l=l;
 53     e[rt].r=r;
 54     e[rt].lazy=-1;
 55     if(l==r) {
 56         e[rt].sum=0;
 57         return;
 58     }
 59     int mid=mi();
 60     build(lson);
 61     build(rson);
 62     pushdown(rt);
 63 }
 64 
 65 int query(int l,int r,int rt) {
 66     if(e[rt].l==l&&e[rt].r==r) {
 67         return e[rt].sum;
 68     }
 69     pushup(rt);
 70     int mid=(e[rt].l+e[rt].r)>>1;
 71     if(r<=mid)  return query(l,r,rt<<1);
 72     else if(l>mid) return query(l,r,rt<<1|1);
 73     else return query(l,mid,rt<<1)+query(mid+1,r,rt<<1|1);
 74 }
 75 
 76 void update(int l,int r,int rt,int c){
 77      if(e[rt].l==l&&e[rt].r==r){
 78          e[rt].lazy=c;
 79          e[rt].sum=(e[rt].r-e[rt].l+1)*c;
 80          return;
 81      }
 82      pushup(rt);
 83      int mid=(e[rt].l+e[rt].r)>>1;
 84      if(r<=mid) update(l,r,rt<<1,c);
 85      else if(l>mid) update(l,r,rt<<1|1,c);
 86      else {
 87            update(l,mid,rt<<1,c);
 88            update(mid+1,r,rt<<1|1,c);
 89     }
 90      pushdown(rt);
 91 }
 92 
 93 int main() {
 94     // fre();
 95     int T;
 96     scanf("%d",&T);
 97     while(T--) {
 98         int n,q;
 99         cin>>n>>q;
100         build(1,n,1);
101         while(q--) {
102             int op,a,b;
103             cin>>op>>a>>b;
104             if(op==1) {
105                 int L=a+1,R=n;
106                 int st,ed;
107                 if((n-L+1-query(L,n,1))==0) {
108                     printf("Can not put any one.\n");
109                     continue;
110                 }
111                 st=inf;
112                 while(L<=R) {
113                     int mid=(L+R)>>1;
114                     if((mid-(a+1)+1-query(a+1,mid,1))>=1) {
115                         st=min(st,mid);
116                         R=mid-1;
117                     } else {
118                         L=mid+1;
119                     }
120                 }
121                 int tem=n-st+1-query(st,n,1);
122                 if(tem<b) b=tem;
123                 ed=inf;
124                 L=st,R=n;
125                 while(L<=R) {
126                     int mid=(L+R)>>1;
127                     tem=mid-st+1-query(st,mid,1);
128                     if(tem==b) {
129                         ed=min(ed,mid);
130                         R=mid-1;
131                     } else if(tem>b) {
132                         R=mid-1;
133                     } else {
134                         L=mid+1;
135                     }
136                 }
137                 printf("%d %d\n",st-1,ed-1);
138                 update(st,ed,1,1);
139             } else {
140                  printf("%d\n",query(a+1,b+1,1));
141                  update(a+1,b+1,1,0);
142             }
143         }
144         cout<<endl;
145     }
146     return 0;
147 }

 

HDU4614 Vases and Flowers

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原文地址:http://www.cnblogs.com/ITUPC/p/5638659.html

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