第九周 4.25 --- 5.1

新的一周

话说这周开始考试了....

4.25

昨天去西电的校赛

过了一年，又去签到了....O_<

 

uestc的数据结构专题

A 卿学姐与公主

最基础的线段树的单点更新

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 typedef long long LL;
 8 const int maxn = 4e6+5;
 9 int n,m,q;
10 LL maxx[maxn],a[maxn];
11 int ql,qr,pos,v;
12 
13 void pushup(int o){
14     maxx[o] = max(maxx[o<<1],maxx[o<<1|1]);
15 }
16 
17 void Build(int o,int l,int r){
18     if(l == r){
19         maxx[o] = 0;
20         return;
21     }
22     int mid = (l+r)/2;
23     Build(o<<1,l,mid);
24     Build(o<<1|1,mid+1,r);
25     pushup(o);
26 }
27 
28 void Update(int o,int l,int r){
29     if(l == r){
30         maxx[o] += 1LL*v;
31         return;
32     }
33     int mid = (l+r)/2;
34     if(pos <= mid) Update(o<<1,l,mid);
35     else Update(o<<1|1,mid+1,r);
36     pushup(o);
37 }
38 
39 LL Query(int o,int l,int r){
40     if(ql <= l && r <= qr){
41         return maxx[o];
42     }
43     int mid = (l+r)/2;
44     LL ans = 0LL;
45     if(ql <= mid) ans = max(ans,Query(o<<1,l,mid));
46     if(qr > mid) ans = max(ans,Query(o<<1|1,mid+1,r));
47     return ans; 
48 }
49 
50 void solve(){
51     Build(1,1,n);
52     int cmd;
53     for(int i = 1;i <= q;i++){
54         scanf("%d",&cmd);
55         if(cmd == 1){
56             scanf("%d %d",&pos,&v);
57             Update(1,1,n);
58         }
59         else{
60             scanf("%d %d",&ql,&qr);
61             LL ans = Query(1,1,n);
62             printf("%lld\n",ans);
63         }
64     }
65 }
66 
67 int main(){
68     while(scanf("%d %d",&n,&q) != EOF){
69         solve();
70     }
71     return 0;
72 }

View Code

 

4.26

N 秋实大哥搞算数

算一个算术表达式的值，不含括号

啊.....不知道栈的经典应用（求算术表达式   TwT......），上来开始敲模拟.....越敲越敲不下去.....

搜了下题解.....又get一点点

先把题目给的中缀表达式 转化成 后缀表达式，再用一个栈去求解

wa 了两发，，是因为，每次碰到一个操作符的时候，需要从栈里面弹两个数出来，但是有时候栈里面不够两个数，就用0来算

还有就是 表达式 最开始 可能 是 -1+5这样

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<stack>
  6 #include<vector>
  7 using namespace std;
  8 
  9 typedef long long LL;
 10 const int maxn = 1e6+5;
 11 char s[maxn],t[maxn];
 12 int n;
 13 LL p[20];
 14 
 15 LL suan(string tmp){
 16     int len = tmp.length();
 17     LL res = 0LL;
 18     for(int i = 0;i < len;i++) {
 19         res += 1LL*(tmp[i]-'0')*p[len-i-1];
 20     }
 21     return res;
 22 }
 23 
 24 int Rank(char c){
 25     if(c == '+' || c == '-') return 1;
 26     if(c == '*' || c == '/') return 2;
 27 }
 28 
 29 vector<string> num;
 30 
 31 void solve(){
 32     num.clear();
 33     n = strlen(t+1);
 34     if(t[1] == '-'){
 35         s[1] = '0';
 36         for(int i = 2;i <= n+1;i++) s[i] = t[i-1];
 37         n++;
 38     }
 39     else{
 40         for(int i = 1;i <= n;i++) s[i] = t[i];
 41     }
 42 
 43     stack<char> ss;
 44     for(int i = 1;i <= n;){
 45         if(s[i] >= '0' && s[i] <= '9'){
 46             int j = i;
 47             string tmp;
 48             while(j <= n && (s[j] >= '0' && s[j] <= '9')){
 49                 tmp += s[j];
 50                 j++;
 51             }
 52             i = j;
 53             num.push_back(tmp);
 54         }
 55         else{
 56             int x = Rank(s[i]);
 57         //    printf("s[%d] = %c\n",i,s[i]);
 58             while(!ss.empty() && x <= Rank(ss.top())){
 59                 string now;
 60                 now += ss.top();
 61                 num.push_back(now);ss.pop();
 62             }
 63             ss.push(s[i]);
 64             i++;
 65         }
 66     }
 67     while(!ss.empty()){
 68         string now;
 69         now += ss.top();
 70         num.push_back(now);ss.pop();
 71     }
 72     //for(int i = 0;i < num.size();i++) printf("%s",num[i].c_str());
 73     //printf("\n");
 74     stack<LL> sss;
 75     for(int i = 0;i < num.size();i++) {
 76         string now = num[i];
 77         if(now != "+" && now != "-" && now != "*" && now != "/"){
 78             LL y = suan(now);
 79             //printf("y = %I64d\n",y);
 80             sss.push(y);
 81         }
 82         else{
 83             LL x,y;
 84             if(sss.empty()) x = 0LL;
 85             else {
 86                 x = sss.top();sss.pop();
 87             }
 88             if(sss.empty()) y = 0LL;
 89             else{
 90                 y = sss.top();sss.pop();
 91             }
 92             
 93             if(now == "+"){
 94                 sss.push(x+y);
 95             }
 96             if(now == "-"){
 97                 sss.push(y-x);
 98             }
 99             if(now == "*"){
100                 sss.push(x*y);
101             }
102             if(now == "/"){
103                 sss.push(y/x);
104             }
105         }
106     }    
107     printf("%lld\n",sss.top());
108 }
109 
110 int main(){
111     int T;
112     p[0] = 1LL;
113     for(int i = 1;i <= 18;i++) p[i] = p[i-1]*10LL;
114     scanf("%d",&T);
115     while(T--){
116         scanf("%s",t+1);
117         solve();
118     }
119     return 0;
120 }

View Code

 

4.27

看了一题，，不会捉....

 

4.28

wawawawawa......

 

4.29

补西电校赛的题

Tom的树 

一神教的，用并查集维护个数，算贡献

话说用I64d wa，，要用 lld

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 typedef long long LL;
 8 const int maxn = 1e5+5;
 9 int sz[maxn],fa[maxn];
10 int n;
11 
12 struct Edge{
13     int u,v,w;
14 }e[maxn];
15 
16 int cmp1(Edge n1,Edge n2){
17     return n1.w < n2.w;
18 }
19 
20 int cmp2(Edge n1,Edge n2){
21     return n1.w > n2.w;
22 }
23 
24 int find(int x){return x == fa[x]? x : fa[x] = find(fa[x]);}
25 
26 void solve(){
27     LL maxx = 0LL;
28     LL minn = 0LL;
29     sort(e+1,e+n,cmp1);
30     for(int i = 1;i <= n;i++) fa[i] = i,sz[i] = 1; 
31     
32     for(int i = 1;i <= n-1;i++){
33         int u = e[i].u;
34         int v = e[i].v;
35         int w = e[i].w;
36         int x = find(u);
37         int y = find(v);
38         LL tmp = 1LL*sz[x]*sz[y]*w;
39         if(x != y){
40             fa[x] = y;
41             sz[y] += sz[x];
42         }
43         maxx += tmp;
44     }
45     //printf("maxx = %I64d\n",maxx);
46 
47     sort(e+1,e+n,cmp2);
48     for(int i = 1;i <= n;i++) fa[i] = i,sz[i] = 1; 
49     
50     for(int i = 1;i <= n-1;i++){
51         int u = e[i].u;
52         int v = e[i].v;
53         int w = e[i].w;
54         int x = find(u);
55         int y = find(v);
56         LL tmp = 1LL*sz[x]*sz[y]*w;
57         if(x != y){
58             fa[x] = y;
59             sz[y] += sz[x];
60         }
61         minn += tmp;
62     }
63     //printf("minn = %I64d\n",minn);
64     LL ans = maxx - minn;
65     printf("%lld\n",ans);
66 
67 }
68 
69 int main(){
70     while(scanf("%d",&n) != EOF){
71         for(int i = 1;i <= n-1;i++){
72             scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].w);
73         }
74         solve();
75     }
76     return 0;
77 }

View Code

 

又好久没做 leetcode 了....

下周开始各种考试了TwT/

 

leetcode 225 Implement Stack using Queues

两个队列实现一个栈

 1 class Stack{
 2 public:
 3     queue<int>q,q2;
 4     void push(int x){
 5         q.push(x);
 6     }
 7 
 8     void pop(){
 9         while(q.size() > 1){
10             q2.push(q.front());
11             q.pop();
12         }
13         q.pop();
14         while(q2.size() > 0){
15             q.push(q2.front());
16             q2.pop();
17         }
18     }
19     
20     int top(){
21         int tt;
22         while(q.size() > 0){
23             tt = q.front();
24             q2.push(q.front());
25             q.pop();
26         }
27         while(q2.size() > 0){
28             q.push(q2.front());
29             q2.pop();
30         }
31         return tt;
32     }
33 
34     bool empty(){
35         return q.empty();
36     }
37 };

View Code

 

4.30

leetcode 155 Min Stack

实现一个能够知道当前最小值的栈

另外再维护一个最小值的栈就好了

 1 class MinStack{
 2 public:
 3     stack<int> s,minn;
 4     void push(int x){
 5         if(s.empty()){
 6             s.push(x);
 7             minn.push(x);
 8         }
 9         else{
10             s.push(x);
11             if(x <= minn.top()) minn.push(x);
12         }
13     }
14 
15     void pop(){
16         if(s.top() == minn.top()){
17             minn.pop();
18         }
19         s.pop();
20     }
21 
22     int top(){
23         return s.top();
24     }
25 
26     int getMin(){
27         return minn.top();
28     }
29 };

View Code

 

5.1

劳动节彩蛋....

bc 爆零....

诶....sbsbsbsbsb

leetcode 110 Balanced Binary Tree

判断一颗二叉树是否是平衡的二叉树

 1 class Solution{
 2 public:
 3     int h(TreeNode* root){
 4         if(root == NULL) return 0;
 5         if(root->left == NULL && root->right == NULL) return 1;
 6         return max(h(root->left),h(root->right))+1;
 7     }
 8 
 9     bool isBalanced(TreeNode* root){
10         if(root == NULL) return true;
11         if(abs(h(root->left) - h(root->right)) > 1) return false;
12         return isBalanced(root->left) && isBalanced(root->right);
13     }
14 };

View Code

 

转载于:https://www.cnblogs.com/wuyuewoniu/p/5432422.html