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
时间： 2024-08-09 02:21:00