洛谷试炼场 提高模板-nlogn数据结构

 

树状数组-区间求和

P3374 【模板】树状数组 1

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 int read(){
 9     int x=0,f=1;char ch=getchar();
10     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
11     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
12     return x*f;
13 }
14 const int mxn=500010;
15 int n,m;
16 int t[mxn];
17 void add(int x,int v){
18     while(x<mxn){t[x]+=v;x+=x&-x;}
19 }
20 int smm(int x){
21     int res=0;
22     while(x){
23         res+=t[x];
24         x-=x&-x;
25     }
26     return res;
27 }
28 int main(){
29     int op,x,k;
30     n=read();m=read();
31     int i,j;
32     for(i=1;i<=n;i++){
33         x=read();
34         add(i,x);
35     }
36     for(i=1;i<=m;i++){
37         op=read();x=read();k=read();
38         if(op==1){
39             add(x,k);
40         }
41         else{
42             printf("%d\n",smm(k)-smm(x-1));
43         }
44     }
45     return 0;
46 }
树状数组1

 

树状数组-差分

P3368 【模板】树状数组 2

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 int read(){
 9     int x=0,f=1;char ch=getchar();
10     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
11     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
12     return x*f;
13 }
14 const int mxn=500010;
15 int n,m;
16 int t[mxn];
17 void add(int x,int v){
18     while(x<mxn){t[x]+=v;x+=x&-x;}
19 }
20 int smm(int x){
21     int res=0;
22     while(x){
23         res+=t[x];
24         x-=x&-x;
25     }
26     return res;
27 }
28 int main(){
29     int op,x,k;
30     n=read();m=read();
31     int i,j;
32     for(i=1;i<=n;i++){
33         x=read();
34         add(i,x);
35         add(i+1,-x);
36     }
37     for(i=1;i<=m;i++){
38         op=read();
39         if(op==1){
40             x=read();k=read();op=read();
41             add(x,op);
42             add(k+1,-op);
43         }
44         else{
45             x=read();
46             printf("%d\n",smm(x));
47         }
48     }
49     return 0;
50 }
树状数组2

 

线段树-区间加 区间乘

P3373 【模板】线段树 2

  1 /*by SilverN*/
  2 #include<iostream>
  3 #include<algorithm>
  4 #include<cstring>
  5 #include<cstdio>
  6 #include<cmath>
  7 #define ls l,mid,rt<<1
  8 #define rs mid+1,r,rt<<1|1
  9 #define lc rt<<1
 10 #define rc rt<<1|1
 11 using namespace std;
 12 const int mxn=1000000;
 13 long long n,p;
 14 long long a[mxn];
 15 struct node{
 16     long long sum;
 17     long long mu;
 18     long long add;
 19 }tr[mxn];
 20 void pushdown(int rt,int m){
 21     tr[lc].sum=(tr[lc].sum*tr[rt].mu+(m-(m>>1))*tr[rt].add)%p;
 22                                 //m-(m>>1)得到区间范围的一半，也就是左子树的范围 
 23     tr[rc].sum=(tr[rc].sum*tr[rt].mu+(m>>1)*tr[rt].add)%p;
 24     tr[lc].mu=tr[lc].mu*tr[rt].mu%p;
 25     tr[rc].mu=tr[rc].mu*tr[rt].mu%p;
 26     tr[lc].add=(tr[lc].add*tr[rt].mu+tr[rt].add)%p;
 27     tr[rc].add=(tr[rc].add*tr[rt].mu+tr[rt].add)%p;
 28     tr[rt].mu=1;tr[rt].add=0;
 29     return;
 30 }
 31 void Build(int l,int r,int rt){
 32     tr[rt].mu=1;tr[rt].add=0;
 33     if(l==r){
 34         tr[rt].sum=a[l];
 35         return;
 36     }
 37     int mid=(l+r)>>1;
 38     Build(ls);
 39     Build(rs);
 40     tr[rt].sum=(tr[rt<<1].sum+tr[rt<<1|1].sum)%p;
 41     return;
 42 }
 43 void add(int L,int R,int l,int r,int rt,int v){
 44     if(L<=l && r<=R){
 45         tr[rt].sum=(tr[rt].sum+v*(r-l+1))%p;//本身值累加区间新增值
 46         tr[rt].add=(tr[rt].add+v)%p;//标记累加 
 47         return; 
 48     }
 49     pushdown(rt,r-l+1);//下传 
 50     int mid=(l+r)>>1;
 51     if(L<=mid)add(L,R,ls,v);
 52     if(R>mid)add(L,R,rs,v);
 53     tr[rt].sum=(tr[lc].sum+tr[rc].sum)%p;
 54     return;
 55 }
 56 void multi(int L,int R,int l,int r,int rt,int v){
 57     if(L<=l && r<=R){
 58         tr[rt].add=(tr[rt].add*v)%p;
 59         tr[rt].mu=(tr[rt].mu*v)%p;
 60         tr[rt].sum=(tr[rt].sum*v)%p;
 61         return;
 62     }
 63     pushdown(rt,r-l+1);
 64     int mid=(l+r)>>1;
 65     if(L<=mid)multi(L,R,ls,v);
 66     if(R>mid)multi(L,R,rs,v);
 67     tr[rt].sum=(tr[lc].sum+tr[rc].sum)%p;
 68     return;
 69 }
 70 long long query(int L,int R,int l,int r,int rt){//查询 
 71     if(L<=l && r<=R)return tr[rt].sum%p;
 72     int mid=(l+r)>>1;
 73     pushdown(rt,r-l+1);
 74     long long res=0;
 75     if(L<=mid)res=(res+query(L,R,ls))%p;
 76     if(R>mid)res=(res+query(L,R,rs))%p;
 77     tr[rt].sum=(tr[lc].sum+tr[rc].sum)%p;
 78     return res%p;
 79 }
 80 int main(){
 81     int M;
 82     scanf("%lld%lld%lld",&n,&M,&p);
 83     int i,j;
 84     for(i=1;i<=n;i++)scanf("%lld",&a[i]);
 85     Build(1,n,1);
 86     int op,g,t,c;
 87     while(M--){
 88         scanf("%d%d%d",&op,&t,&g);
 89         if(op==1){//乘 
 90             scanf("%d",&c);
 91             multi(t,g,1,n,1,c);
 92         }
 93         if(op==2){//加 
 94             scanf("%d",&c);
 95             add(t,g,1,n,1,c);
 96         }
 97         if(op==3){//询问 
 98             long long ans=query(t,g,1,n,1);
 99             printf("%lld\n",ans%p);
100         }
101     }
102     return 0;
103 }
线段树区间修改

 

二叉堆

P3378 【模板】堆

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=1200000;
 9 int read(){
10     int x=0,f=1;char ch=getchar();
11     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 int tp[mxn];
16 int cnt=0;
17 void add(int x){
18     tp[++cnt]=x;
19     int now=cnt;
20     while(now>1){
21         int nxt=now/2;
22         if(tp[nxt]<=tp[now])return;
23         swap(tp[nxt],tp[now]);
24         now=nxt;
25     }
26     return;
27 }
28 int del_head(){
29     int res=tp[1];
30     tp[1]=tp[cnt--];
31     int now=1;
32     while(now<=cnt/2){
33         int nxt=now<<1;
34         if(nxt<cnt && tp[nxt+1]<tp[nxt])nxt++;
35         if(tp[now]>tp[nxt]){
36             swap(tp[now],tp[nxt]);
37             now=nxt;
38         }
39         else return res;
40     }
41     return res;
42 }
43 int main(){
44     int n;
45     n=read();
46     int op,x,y;
47     int i,j;
48     for(i=1;i<=n;i++){
49         op=read();
50         switch(op){
51             case 1:{x=read();add(x);break;}
52             case 2:{printf("%d\n",tp[1]);break;}
53             case 3:{del_head();break;}
54         }
55     }
56     return 0;
57 }
最小堆

 

此处是最小堆

posted @ 2016-09-26 18:14  SilverNebula  阅读(397)  评论(0)  编辑  收藏  举报
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