1、树状数组
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <cmath>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8 #define INF ~0U>>1
9
10 const int MAX=300;
11 int n,num[MAX],c[MAX],X,Y;
12
13 int lowbit(int x){ return x&(-x);}
14
15 int SUM(int l,int r){//区间查询
16 int sum=0;
17
18 for(int i=r;i;i-=lowbit(i)) sum+=c[i];
19 for(int i=l;i;i-=lowbit(i)) sum-=c[i];
20
21 return sum;
22 }
23
24 void change(int x,int m){ for(int i=x;i<=n;i+=lowbit(i)) c[i]+=m;}//单点修改
25
26 int main(){
27 memset(c,0,sizeof(c));
28
29 cin >> n;
30 for(int i=1;i<=n;i++){
31 cin >> num[i];
32 for(int j=i-lowbit(i)+1;j<=i;j++) c[i]+=num[j];
33 }
34
35 cout << "1 X Y:sum[Y]-sum[X](X<Y)" << endl << "2 X Y:num[X]=Y" << endl << "0:end" << endl;
36
37 int i;
38 do{
39 cin >> i >> X >> Y;
40 if(!i) break;
41 if(i==1) cout << SUM(X,Y) << endl;
42 if(i==2) change(X,Y),cout << "FINISH!" << endl;
43 }while(i);
44
45 return 0;
46 }
2、线段树
http://blog.csdn.net/metalseed/article/details/8039326
①单点修改替换，查询区间最值区间求和
1 #include<iostream>
2 #include<algorithm>
3 #define lson l,m,rt<<1
4 #define rson m+1,r,rt<<1|1
5 #define maxn 10000
6 int n,rt,Max[maxn<<2],Min[maxn<<2],Sum[maxn<<2],a[maxn<<2];
7 using namespace std;
8 void pushup(int rt){
9 Max[rt] = max(Max[rt<<1],Max[rt<<1|1]);
10 Min[rt] = min(Min[rt<<1],Min[rt<<1|1]);
11 Sum[rt] = Sum[rt<<1]+Sum[rt<<1|1];
12 }
13 void buildtree(int l,int r,int rt){
14 if(l == r){
15 Max[rt] = Min[rt] = Sum[rt] = a[l];
16 return;
17 }
18 int m = (l+r)>>1;
19 buildtree(lson);
20 buildtree(rson);
21 pushup(rt);
22 }
23 void setval(int p,int v,int l,int r,int rt){
24 if(l == r){
25 Max[rt] = Min[rt] = Sum[rt] = v;
26 return;
27 }
28 int m = (l+r)>>1;
29 if(p <= m) update(p,v,lson);
30 else update(p,v,rson);
31 pushup(rt);
32 }
33 void addval(int p,int v,int l,int r,int rt){
34 if(l == r){
35 Max[rt] += v;
36 Min[rt] += v;
37 Sum[rt] += v;
38 return;
39 }
40 int m = (l+r)>>1;
41 if(p <= m) update(p,v,lson);
42 else update(p,v,rson);
43 pushup(rt);
44 }
45
46 int query_max(int L,int R,int l,int r,int rt){
47 if(L <= l && r <= R) return Max[rt];
48 int m = (l+r)>>1;
49 int ret = 0;
50 if(L <= m) ret = max(ret,query(L,R,lson));
51 if(R > m) ret = max(ret,query(L,R,rson));
52 return ret;
53 }
54 int query_min(int L,int R,int l,int r,int rt){
55 if(L <= l && r <= R) return Min[rt];
56 int m = (l+r)>>1;
57 int ret = 0;
58 if(L <= m) ret = min(ret,query(L,R,lson));
59 if(R > m) ret = min(ret,query(L,R,rson));
60 return ret;
61 }
62 int query_sum(int L,int R,int l,int r,int rt){
63 if(L <= l && r <= R) return Sum[rt];
64 int m = (l+r)>>1;
65 int ret = 0;
66 if(L <= m) ret += query(L,R,lson);
67 if(R > m) ret += query(L,R,rson);
68 return ret;
69 }
70
71 int main(){
72 cin>>n;
73 buildtree(1,n,1);
74 return 0;
75 }
②区间修改，查询区间最值区间求和
1 #include<iostream>
2
3 #include<algorithm>
4
5 using namespace std;
6
7 const long long int INF = 100000000;
8
9 const int maxn = 2000000 + 10;
10
11 long long int sumv[maxn], minv[maxn], maxv[maxn], addv[maxn];//sumv[]：如果只执行结点o及其子孙结点中的add操作，结点o对应区间中所有数之和
12
13 long long int my_y1, my_y2, v, n, m;//修改/查询范围均为[my_y1,my_y2];
14
15 void mantain(int o, int l, int r){
16
17 int lc = o * 2, rc = o * 2 + 1;
18
19 sumv[o] = minv[o] = maxv[o] = 0;
20
21 if (r > l){//考虑左右子树
22
23 sumv[o] = sumv[lc] + sumv[rc];
24
25 minv[o] = min(minv[lc], minv[rc]);
26
27 maxv[o] = max(maxv[lc], maxv[rc]);
28
29 }
30
31 minv[o] += addv[o]; maxv[o] += addv[o]; sumv[o] += addv[o] * (r - l + 1);
32
33 //考虑add操作
34
35 }
36
37 //在执行add操作时，哪些结点需啊哟调用上述maintain函数呢？很简单，递归访问到的结点全部要调用，并且是在递归返回后调用。
38
39 void update(int o, int l, int r){
40
41 int lc = o * 2, rc = o * 2 + 1;
42
43 if (my_y1 <= l&&my_y2 >= r){//递归边界
44
45 addv[o] += v;
46
47 }
48
49 else{
50
51 int m = l + (r - l) / 2;
52
53 if (my_y1 <= m) update(lc, l, m);
54
55 if (my_y2 > m) update(rc, m + 1, r);
56
57 }
58
59 mantain(o, l, r);
60
61 }
62
63
64
65 long long int _min, _max, _sum;//全局变量，目前位置的最小值、最大值和累加和
66
67 void query(int o, int l, int r, int add){
68
69 if (my_y1 <= l&&my_y2 >= r){//递归边界：用边界区间的附加信息更新答案
70
71 _sum += sumv[o] + add*(r - l + 1);
72
73 _min = min(_min, minv[o] + add);
74
75 _max = max(_max, maxv[o] + add);
76
77 }
78
79 else{//递归统计，累加参数add
80
81 int m = l + (r - l) / 2;
82
83 if (my_y1 <= m) query(o * 2, l, m, add + addv[o]);
84
85 if (my_y2 > m) query(o * 2 + 1, m + 1, r, add + addv[o]);
86
87
88
89 }
90
91 }
92
93 int main(){
94
95 int i, j;
96
97 cin >> n;
98
99 for (i = 1; i <= n; i++)
100
101 {
102
103 cin >> v;
104
105 my_y1 = my_y2 = i;
106
107 update(1, 1, n);
108
109 }
110
111 cin >> m;
112
113 for (i = 1; i <= m; i++){
114
115 int q;
116
117 cin >> q;
118
119 if (q == 1){
120
121 cin >> my_y1 >> my_y2 >> v;
122
123 update(1, 1, n);
124
125
126
127 }
128
129 else{
130
131 int y;
132
133 cin >> my_y1>>my_y2;
134
135 _sum = 0; _min = INF; _max = -INF;
136
137 query(1, 1, n, 0);
138
139 cout << _sum << endl;
140
141 }
142
143 }
144
145 system("pause");
146
147 }
③区间替换，区间求和
1 #include<iostream>
2 #include<algorithm>
3 #define lson l,m,rt<<1
4 #define rson m+1,r,rt<<1|1
5 #define maxn 10000
6 int n,rt,sum[maxn<<2],col[maxn<<2];//col是延迟标记
7 using namespace std;
8 void pushup(int rt){
9 sum[rt] = sum[rt<<1] + sum[rt<<1|1];//或min,max
10 }
11 void pushdown(int rt,int m){
12 if(col[rt]){
13 col[rt<<1] = col[rt<<1|1] = col[rt];
14 sum[rt<<1] = (m-(m>>1)) * col[rt];
15 sum[rt<<1|1] = (m>>1) * col[rt];
16 col[rt] = 0;
17 }
18 }
19 void buildtree(int l,int r,int rt){
20 col[rt] = 0;
21 sum[rt] = 1;
22 if(l == r) return;
23 int m = (l+r)>>1;
24 buildtree(lson);
25 buildtree(rson);
26 pushup(rt);
27 }
28 void update(int L,int R,int c,int l,int r,int rt){
29 if(L <= l && r <= R){
30 col[rt] = c;
31 sum[rt] = c*(r-l+1);
32 return;
33 }
34 pushdown(rt,r-l+1);
35 int m = (l+r)>>1;
36 if(L <= m) update(L,R,c,lson);
37 if(R>m) update(L,R,c,rson);
38 pushup(rt);
39 }
40 int query(int L,int R,int l,int r,int rt){
41 if(L <= l && r <= R) return sum[rt];
42 int m = (l+r)>>1;
43 int ret = 0;
44 if(L <= m) ret += query(L,R,lson);
45 if(R > m) ret += query(L,R,rson);
46 return ret;
47 }
48 int main(){
49 cin>>n;//5
50 buildtree(1,n,1);//12345
51 update(1,5,2,1,n,1);
52 update(5,9,3,1,n,1);
53 cout<<query(1,n,1,n,1);
54 return 0;
55 }
3、Spare Table
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4 #include <algorithm>
5 #include <cstdlib>
6 #include <cstring>
7 using namespace std;
8 #define MAX 50010
9
10 int top,n,x,y;
11 int f[MAX*2][30],f2[MAX*2][30];
12 int seq[MAX*2];
13
14 int rmq(){
15 int m=(int)(log((double)top-1)/log(2.0));
16
17 for(int i=1;i<top;++i) f[i][0]=seq[i],f2[i][0]=seq[i];
18
19 for(int j=1,k=1<<(j-1);j<=m;++j,k=1<<(j-1))
20 for(int i=1;i+(1<<(j-1))<top;++i){
21 f[i][j]=min(f[i][j-1],f[i+k][j-1]);
22 f2[i][j]=max(f2[i][j-1],f2[i+k][j-1]);
23 }
24
25 return 0;
26 }
27
28 int query(int l,int r){
29 int k=(int)(log(r-l+1.0)/log(2.0));
30
31 if(l>r)swap(l,r);
32
33 return max(f2[l][k],f2[r-(1<<k)+1][k])-min(f[l][k],f[r-(1<<k)+1][k]);
34 }
35
36 int main(){
37 cin >> top >> n;
38 top++;
39
40 for(int i=1;i<top;i++) cin >> seq[i];
41
42 rmq();
43
44 while(n--){
45 cin >> x >> y;
46 cout << query(x,y) << endl;
47 }
48
49 return 0;
50 }
时间： 2024-10-14 14:31:01