0x17 二叉堆

最新推荐文章于 2024-08-07 10:08:16 发布

转载最新推荐文章于 2024-08-07 10:08:16 发布 · 80 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/kkkstra/p/11118506.html

本文详细介绍了手写模板实现的堆操作，包括插入、查找最大值、删除等，并利用Huffman树解决最小化加权路径长度问题。通过具体例题展示了不同场景下算法的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

手写模板：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <stack>
 7 using namespace std;
 8 const int INF=0x3f3f3f3f;
 9 const int maxn=100000+10;
10 int n;
11 int heap[maxn];
12 void up(int p) {
13     while (p>1) {
14         if (heap[p]>heap[p/2]) {
15             swap(heap[p], heap[p/2]);
16             p/=2;
17         }
18         else break;
19     }
20 }
21 void insert(int val) {
22     heap[++n]=val;
23     up(n);
24 }
25 int top() {
26     return heap[1];
27 }
28 void down(int p) {
29     int s=p*2;
30     while (s<=n) {
31         if (s<n&&heap[s]<heap[s+1]) s++;    //左右子节点取较大者
32         if (heap[p]<heap[s]) {
33             swap(heap[p], heap[s]);
34             p=s; s=p*2;
35         }
36         else break;
37     } 
38 }
39 void extract() {
40     heap[1]=heap[n--];
41     down(1);
42 }
43 void remove(int k) {
44     heap[k]=heap[n--];
45     up(k), down(k);
46 }
47 
48 int main() {
49     //freopen("a.txt", "r", stdin);
50     //freopen("a.out", "w", stdout);
51     scanf("%d", &n);
52     for (int i=1; i<=n; ++i) {
53         int x;
54         scanf("%d", &x);
55         insert(x);
56         printf("%d\n", top());
57     }
58     return 0;
59 }

View Code

STL：

#include <queue>
priority_queue<int> q;

【例题】POJ1456 Supermarket

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <queue>
 7 using namespace std;
 8 const int INF=0x3f3f3f3f;
 9 const int maxn=100000+10;
10 int n;
11 struct node {
12     int p, d;
13 } a[maxn];
14 priority_queue<int, vector<int>, greater<int> > Q;
15 
16 bool cmp(node a, node b) {
17     return a.d<b.d;
18 }
19 
20 int main() {
21     //freopen("a.txt", "r", stdin);
22     //freopen("a.out", "w", stdout);
23     while (~scanf("%d", &n)) {
24         for (int i=1; i<=n; ++i)
25             scanf("%d%d", &a[i].p, &a[i].d);
26         sort(a+1, a+n+1, cmp);
27         Q.push(a[1].p);
28         for (int i=2; i<=n; ++i) {
29             int w=Q.top(), sz=Q.size();
30             if (a[i].d==sz && a[i].p>w) {
31                 Q.pop();
32                 Q.push(a[i].p);
33             }
34             else if (a[i].d>sz) {
35                 Q.push(a[i].p);
36             }
37         }
38         int ans=0;
39         while (!Q.empty()) {
40             ans+=Q.top();
41             Q.pop();
42         }
43         printf("%d\n", ans);
44     }
45     return 0;
46 }

View Code

【例题】POJ2442 Sequence

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <queue>
 7 using namespace std;
 8 const int INF=0x3f3f3f3f;
 9 const int maxn=2000+10;
10 const int maxm=100+10;
11 int n, m;
12 int A[maxm][maxn], f[maxn];
13 
14 struct node {
15     int p1, p2, i, j;
16     bool flag;
17     node() {};
18     node(int a, int b, int c, int d, int f) {
19         p1=a, p2=b, i=c, j=d, flag=f;
20     }
21 };
22 bool operator < (const node& a, const node& b) {
23     return (a.i+a.j)>(b.i+b.j);
24 }
25 
26 priority_queue <node> Q;
27 
28 void merge(int k) {
29     while (!Q.empty()) Q.pop();
30     Q.push(node(1, 1, f[1], A[k][1], 0));
31     int c[maxn], cnt=0;
32     while (!Q.empty()) {
33         node x=Q.top(); Q.pop();
34         c[++cnt]=x.i+x.j;
35         if (x.p2<m) Q.push(node(x.p1, x.p2+1, f[x.p1], A[k][x.p2+1], 1));
36         if (!x.flag && x.p1<m) Q.push(node(x.p1+1, x.p2, f[x.p1+1], A[k][x.p2], 0));
37         if (cnt==m) break;
38     }
39     for (int i=1; i<=m; ++i) f[i]=c[i];
40 }
41 
42 int main() {
43     //freopen("a.txt", "r", stdin);
44     //freopen("a.out", "w", stdout);
45     int T;
46     scanf("%d", &T);
47     while (T--) {
48         while (!Q.empty()) Q.pop();
49         scanf("%d%d", &n, &m);
50         for (int i=1; i<=n; ++i)
51             for (int j=1; j<=m; ++j)
52                 scanf("%d", &A[i][j]);
53         for (int i=1; i<=n; ++i) sort(A[i]+1, A[i]+m+1);
54         for (int i=1; i<=m; ++i) f[i]=A[1][i];
55         for (int i=2; i<=n; ++i) merge(i);
56         for (int i=1; i<=m; ++i) printf("%d ", f[i]);
57         puts("");
58     }
59     return 0;
60 }

View Code

Huffman树

构造一棵包含n个叶子节点的k叉树，其中第i个叶子节点带有权值Wi，要求最小化sigma(Wi*Li)。

【例题】CH1701/NOIP2004 合并果子

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <queue>
 7 using namespace std;
 8 const int INF=0x3f3f3f3f;
 9 const int maxn=2000+10;
10 const int maxm=100+10;
11 int n;
12 priority_queue<int, vector<int>, greater<int> > Q;
13 
14 int main() {
15     //freopen("a.txt", "r", stdin);
16     //freopen("a.out", "w", stdout);
17     scanf("%d", &n);
18     for (int i=1; i<=n; ++i) {
19         int x;
20         scanf("%d", &x);
21         Q.push(x);
22     }
23     int ans=0;
24     while (Q.size()!=1) {
25         int x=Q.top(); Q.pop();
26         int y=Q.top(); Q.pop();
27         ans+=x+y;
28         Q.push(x+y);
29     }
30     printf("%d\n", ans);
31     return 0;
32 }