div3 1006 E dfs系列
我的解法：
中规中矩但wa+timelimit了，估计是算法效率太低。
！！！！这个序列只需要递归一遍，从根节点开始，然后根据输入的查询查找序列即可
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 const int maxn=200001; 5 vector<int> que[maxn],ans;//记录每个节点的孩子结点 6 int u,k,n,q,x,cnt; 7 bool dfs(int index) 8 { 9 // cnt:搜索到孩子个数 10 if (cnt==k) 11 { 12 ans.push_back(index); return true; 13 } 14 if (index>n) return false; 15 // cout <<que[index].size()<<"11"<<endl; 16 for (int i=0;i<que[index].size();i++) // 遍历孩子结点 17 { 18 cnt++; 19 // cout<<"index:"<<index<<",cnt:"<<cnt<<",que[index][i]"<<que[index][i]<<endl; 20 if (dfs(que[index][i])) return true; 21 } 22 23 } 24 int main() 25 { 26 while (scanf("%d%d",&n,&q)!=EOF) 27 { 28 for (int i=1;i<=n;i++) que[i].clear(); 29 for (int i=2;i<=n;i++) 30 { 31 cin>>x; 32 que[x].push_back(i); 33 } 34 ans.clear(); 35 while (q--) 36 { 37 cin>>u>>k;//从u点为根开始先序遍历的第k个结点的序号，深搜 38 cnt=1; 39 if (!dfs(u)) ans.push_back(-1); 40 } 41 for (int i=0;i<ans.size();i++) cout << ans[i] << ' '; 42 } 43 44 return 0; 45 }
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int v1[200000 + 1]; 6 int v2[200000 + 1]; 7 int v3[200000 + 1]; 8 vector <int> adjList[200000 + 1]; 9 int cont; 10 11 int dfs(int nodo) { 12 v1[cont] = nodo; 13 ++cont; 14 v3[nodo] = cont; 15 16 int suma = 0; 17 for(int i = 0; i < adjList[nodo].size(); i++) { 18 suma += dfs(adjList[nodo][i]); 19 } 20 21 v2[nodo] += suma+1; 22 return v2[nodo]; 23 } 24 25 int main() { 26 27 int n,q; 28 scanf("%d%d", &n, &q); 29 30 int aux; 31 for (int i = 1; i < n; i++) { 32 scanf("%d", &aux); 33 adjList[aux-1].push_back(i); 34 } 35 36 cont = 0; 37 dfs(0); 38 39 int a,b; 40 while (q--) { 41 scanf("%d%d",&a, &b); 42 if (v2[a-1] >= b) 43 cout << v1[v3[a-1]-2+b]+1 << endl; 44 else 45 cout << -1 << endl; 46 } 47 48 return 0; 49 } 50 #include<bits/stdc++.h> 51 52 using namespace std; 53 54 const int maxn=200001; 55 56 vector<int> que[maxn]; 57 int child[maxn],order[maxn],ans[maxn];//记录每个节点的孩子结点 58 int u,k,n,q,x,cnt; 59 60 int dfs(int index) 61 { 62 ans[cnt++]=index;//依次存放以0为根的序列 63 order[index]=cnt; // index在最终的序列中为第几个位置 64 65 int sum=0; 66 for (int i=0;i<que[index].size();i++) // 遍历孩子结点 67 { 68 // cout<<"index:"<<index<<",que[index][i]："<<que[index][i]<<endl; 69 sum+=dfs(que[index][i]); 70 } 71 child[index]=sum+1; 72 return child[index];//以index为根节点的子节点总个数 73 } 74 int main() 75 { 76 while (scanf("%d%d",&n,&q)!=EOF) 77 { 78 /* 全局变量会默认地初始化为0 而局部变量不会 有内存就是任性 79 for (int i=1;i<=n;i++) que[i].clear(); 80 ans.clear(); 81 memset(child,0,sizeof(child)); 82 memset(order,0,sizeof(order)); 83 */ 84 for (int i=2;i<=n;i++) 85 { 86 cin>>x; 87 que[x].push_back(i); 88 } 89 90 cnt=0; 91 dfs(1); 92 // for (int i=0;i<ans.size();i++) cout << ans[i] <<" "; cout << endl; 93 94 while (q--) 95 { 96 cin>>u>>k;//从u点为根开始先序遍历的第k个结点的序号，深搜 97 if (child[u]>=k) 98 { 99 cout << ans[order[u]+k-2] << endl; 100 } 101 else 102 cout << -1 << endl; 103 } 104 } 105 return 0; 106 }
A. New Year Transportation
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, …, an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can’t leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don’t know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, …, an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print “YES”. Otherwise, print “NO”.
Examples
inputCopy
8 4
1 2 1 2 1 2 1
outputCopy
YES
inputCopy
8 5
1 2 1 2 1 1 1
outputCopy
NO1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn=30005; 6 int n,t,a[maxn],j; 7 8 int main() 9 { 10 int flag; 11 while (scanf("%d%d",&n,&t)!=EOF) 12 { 13 flag=false; 14 for (int i=1;i<n;i++) { 15 scanf("%d",&a[i]); 16 a[i]+=i; 17 } 18 for (j=1;j<t;j=a[j]); 19 // 不停地往下跳，j是一定小于t的 20 cout << (j==t?"YES":"NO")<<endl; 21 } 22 return 0; 23 }
没有理解清楚题意，很巧妙的解法。
B. BerSU Ball
outputstandard output
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners’ dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, …, an (1 ≤ ai ≤ 100), where ai is the i-th boy’s dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, …, bm (1 ≤ bj ≤ 100), where bj is the j-th girl’s dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
inputCopy
4
1 4 6 2
5
5 1 5 7 9
outputCopy
31 #include<bits/stdc++.h> 2 #define f(s,a) for(int i=0;i<s;i++) cin>>a[i]; 3 // 避免代码重复度高，只需要替换两个字符便能解决的函数 4 using namespace std; 5 6 int n,m; 7 8 int main() 9 { 10 int n,m,c,a[1000],b[1000]; 11 cin>>n; 12 f(n,a) 13 cin>>m; 14 f(m,b) 15 sort(a,a+n); 16 sort(b,b+m); 17 for(int i=0,j=0;i<n && j<m;) // for循环的循环增长条件可以不写 18 if(a[i]-1>b[j]) j++; 19 else 20 if(a[i]<b[j]-1) i++; 21 else c++,i++,j++; 22 // a[i] b[j]的三种关系：a[i]远大于b，b远大于a,以及绝对值相差1以内 23 cout<<c; 24 25 return 0; 26 }
看别人的代码真的能够学到好多东西！！
dfs hdu 搜索练习！
temple of bone hdu1010
明明和正确答案敲得一样，但是却总是WA，气得我一直不停地提交不知道提交了几十遍了。但最后看起来最不起眼的一个地方却是致命点，其实还是我对这个程序的理解不够深！不然怎么会犯这样的错误！全局变量和局部变量的问题+递归调用对局部变量的使用问题！归根结底还是没有研究明白啊你！先读别人的代码！想一想这样写的好处是什么？以及你之前写dfs的时候的问题！
心情舒畅！
我的正确代码：
1 #include <iostream> 2 #include <stdio.h> 3 using namespace std; 4 /* 5 1. 起点终点位置不固定，需要先求出来 6 2. 坐标和的奇偶性：每走一步奇偶性会产生变化 7 */ 8 int n,m,t; 9 char a[10][10]; 10 int path[4][2]= 11 { 12 0,1,1,0,0,-1,-1,0 13 }; 14 bool success; 15 void dfs(int ct,int i,int j) 16 { 17 for (int k=0;k<4;k++) 18 { 19 int ni=i+path[k][0]; 20 int nj=j+path[k][1]; 21 if (ni>=n||nj>=m||ni<0||nj<0) continue; 22 if (a[ni][nj]=='X') continue; 23 if (a[ni][nj]=='D') 24 { 25 if (ct+1==t) 26 { 27 success=true; 28 return; 29 }else 30 continue; 31 } 32 33 a[ni][nj]='X'; 34 dfs(ct+1,ni,nj); 35 a[ni][nj]='.'; 36 if (success) return; 37 } 38 } 39 int main () 40 { 41 int dx,dy; 42 while (scanf("%d%d%d",&n,&m,&t)!=EOF){ 43 if (n==0&&m==0&&t==0) break; 44 success = false; 45 for (int i=0;i<n;i++) scanf("%s",a[i]); 46 for (int i=0;i<n;i++) 47 { 48 for (int j=0;j<m;j++) 49 { 50 if (a[i][j]=='D') {dx=i;dy=j;} 51 } 52 } 53 for (int i=0;i<n;i++) 54 { 55 for (int j=0;j<m;j++) 56 { 57 if (a[i][j]=='S' && (i+j)%2==((dx+dy)%2+t%2)%2){ 58 a[i][j]='X'; 59 dfs(0,i,j); 60 } 61 } 62 } 63 puts(success==true?"YES":"NO"); 64 65 } 66 return 0; 67 } 68 69 另一种我更喜欢写但效率上偏低的写法： 70 #include <iostream> 71 #include <stdio.h> 72 using namespace std; 73 /* 74 1. 起点终点位置不固定，需要先求出来 75 2. 坐标和的奇偶性：每走一步奇偶性会产生变化 76 */ 77 int n,m,t; 78 char a[10][10]; 79 int path[4][2]= 80 { 81 0,1,1,0,0,-1,-1,0 82 }; 83 bool success; 84 void dfs(int ct,int i,int j) 85 { 86 if (a[i][j]=='D') 87 { 88 if (ct==t) 89 success=true; 90 return; 91 } 92 if (i>=n||j>=m||i<0||j<0) return; 93 if (a[i][j]=='X') return; 94 95 for (int k=0;k<4;k++) 96 { 97 a[i][j]='X'; 98 dfs(ct+1,i+path[k][0],j+path[k][1]); 99 a[i][j]='.'; 100 if (success) return; 101 } 102 } 103 int main () 104 { 105 int dx,dy; 106 while (scanf("%d%d%d",&n,&m,&t)!=EOF){ 107 if (n==0&&m==0&&t==0) break; 108 success = false; 109 for (int i=0;i<n;i++) scanf("%s",a[i]); 110 for (int i=0;i<n;i++) 111 { 112 for (int j=0;j<m;j++) 113 { 114 if (a[i][j]=='D') {dx=i;dy=j;} 115 } 116 } 117 for (int i=0;i<n;i++) 118 { 119 for (int j=0;j<m;j++) 120 { 121 if (a[i][j]=='S' && (i+j)%2==((dx+dy)%2+t%2)%2){ 122 dfs(0,i,j); 123 } 124 } 125 } 126 puts(success==true?"YES":"NO"); 127 128 } 129 return 0; 130 }
注意：
- 标注为石头的当前层在递归结束返回该层后要记得还原，否则再往上返回时的棋局和之前便不一样了。
- success的目标是在找到正确答案后立即返回，不进行接下来的循环。
- 或者改为return bool的形式
1 #include <iostream> 2 #include <stdio.h> 3 using namespace std; 4 /* 5 1. 起点终点位置不固定，需要先求出来 6 2. 坐标和的奇偶性：每走一步奇偶性会产生变化 7 */ 8 int n,m,t; 9 char a[10][10]; 10 int path[4][2]= 11 { 12 0,1,1,0,0,-1,-1,0 13 }; 14 bool success; 15 bool dfs(int ct,int i,int j) 16 { 17 if (a[i][j]=='D') 18 { 19 if (ct==t) 20 return true; 21 else return false; 22 } 23 if (i>=n||j>=m||i<0||j<0) return false; 24 if (a[i][j]=='X') return false; 25 26 for (int k=0;k<4;k++) 27 { 28 a[i][j]='X'; 29 success = dfs(ct+1,i+path[k][0],j+path[k][1]); 30 a[i][j]='.'; 31 if (success) return true; 32 } 33 } 34 int main () 35 { 36 int dx,dy; 37 while (scanf("%d%d%d",&n,&m,&t)!=EOF){ 38 if (n==0&&m==0&&t==0) break; 39 success = false; 40 for (int i=0;i<n;i++) scanf("%s",a[i]); 41 for (int i=0;i<n;i++) 42 { 43 for (int j=0;j<m;j++) 44 { 45 if (a[i][j]=='D') {dx=i;dy=j;} 46 } 47 } 48 for (int i=0;i<n;i++) 49 { 50 for (int j=0;j<m;j++) 51 { 52 if (a[i][j]=='S' && (i+j)%2==((dx+dy)%2+t%2)%2){ 53 dfs(0,i,j); 54 } 55 } 56 } 57 puts(success==true?"YES":"NO"); 58 59 } 60 return 0; 61 }
hdu 1016 prime ring problem
1 #include <iostream> 2 #include <stdio.h> 3 #include <vector> 4 #include <string> 5 #include <string.h> 6 7 using namespace std; 8 int n; 9 int a[21]; 10 bool prime[41];// 1-40是所有加和可能出现的素数 11 vector<int> ans; 12 bool visited[21]; 13 void init() 14 { 15 memset(prime,false,sizeof(prime)); 16 for (int i=2;i<=41;i++) 17 { 18 if (prime[i] == true) continue;// 非素数为true 19 for (int j=i*i;j<=41;j+=i) 20 prime[j]=true; 21 } 22 } 23 void print(int num) 24 { 25 // 检查最后一个数和1是否组成素数 26 int i; 27 if (!prime[1+num]) 28 { 29 for (i=0;i<ans.size()-1;i++) cout << ans[i] << " "; 30 cout << ans[i] <<endl; 31 } 32 return; 33 } 34 void dfs(int num,int cnt) 35 { 36 if (cnt==n) 37 { 38 print(num); 39 return; 40 } 41 // num是当前环上的数字，cnt 是1-n个数已经安顿了几个？ 42 for (int i=2;i<=n;i++) 43 { 44 if (num==i) continue; // 当前数字 45 if (visited[i]) continue; // 已经使用过的数字 46 if (prime[i+num]) continue;//和为非素数 47 ans.push_back(i); 48 visited[i]=true; 49 dfs(i,cnt+1); 50 visited[i]=false; 51 ans.pop_back(); 52 } 53 } 54 int main() 55 { 56 int casei=0; 57 init();// prime中2-41间的素数均标记为false 58 while (cin>>n) 59 { 60 ans.clear(); 61 printf("Case %d: ",++casei); 62 memset(visited,false,sizeof(visited)); 63 ans.push_back(1); 64 visited[1]=true; 65 dfs(1,1); 66 printf(" "); 67 } 68 return 0; 69 }
- 素数筛代码要熟练写出
Oil Deposits
Problem Description
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.
Input
The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either *', representing the absence of oil, or@’, representing an oil pocket.
Output
For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
Sample Input
1 1
*
3 5
@@*
@
@@*
1 8
@**@*
5 5
**@
@@@
@*@
@@*@
@**@
0 0
Sample Output
0
1
2
21 #include <iostream> 2 #include <stdio.h> 3 #include <vector> 4 #include <string> 5 #include <string.h> 6 /* 7 相连的@代表一片油田，求油田总个数 8 9 */ 10 using namespace std; 11 int m,n,cnt; 12 char maze[101][101]; 13 int go[8][2]= 14 { 15 -1,0, 16 -1,1, 17 0,1, 18 1,1, 19 1,0, 20 1,-1, 21 0,-1, 22 -1,-1 23 }; 24 void dfs(int x,int y) 25 { 26 maze[x][y]='*'; 27 for (int i=0;i<8;i++) 28 { 29 int newx = go[i][0]+x; 30 int newy = go[i][1]+y; 31 if (maze[newx][newy]!='@') 32 continue; 33 if (newx<0||newy<0||newx>m||newy>n) 34 continue; 35 dfs(newx,newy); 36 } 37 38 } 39 int main() 40 { 41 42 while (cin>>m>>n) 43 { 44 if (m==0&&n==0) break; 45 for (int i=0;i<m;i++) 46 { 47 for (int j=0;j<n;j++) 48 { 49 cin>>maze[i][j]; 50 } 51 } 52 cnt=0; 53 for (int i=0;i<m;i++) 54 { 55 for (int j=0;j<n;j++) 56 { 57 if (maze[i][j]=='@') 58 { 59 dfs(i,j); 60 cnt++; 61 } 62 63 } 64 } 65 cout << cnt << endl; 66 } 67 return 0; 68 }
rescue
求最短的路径，可以找到深搜所有的路径并求最小值，但更好地方法是用bfs求解。
dfs的做法，本地OK但超时。
1 #include <iostream> 2 #include <stdio.h> 3 #include <vector> 4 #include <string> 5 #include <string.h> 6 /* 7 如何求最短的路径？？ 8 9 */ 10 using namespace std; 11 int m,n,mtime; 12 char maze[201][201]; 13 int go[4][2]= 14 { 15 -1,0, 16 0,1, 17 1,0, 18 0,-1 19 }; 20 struct node 21 { 22 int x,y,time; 23 }a,r; 24 void dfs(node cur) 25 { 26 char c=maze[cur.x][cur.y]; 27 for (int i=0;i<4;i++) 28 { 29 node t; 30 t.x=go[i][0]+cur.x; 31 t.y=go[i][1]+cur.y; 32 t.time=cur.time+1; 33 34 if (t.x<0||t.y<0||t.x>m||t.y>n) continue; 35 if (maze[t.x][t.y]=='#') continue; 36 if (maze[t.x][t.y]=='x') {t.time++;maze[t.x][t.y]='.';} 37 if (maze[t.x][t.y]=='a') 38 { 39 a.time=t.time+1; 40 if (a.time<mtime) 41 { 42 mtime=a.time; 43 return; 44 } 45 } 46 maze[cur.x][cur.y]='#'; 47 dfs(t); 48 maze[cur.x][cur.y]=c; 49 } 50 } 51 int main() 52 { 53 54 while (cin>>m>>n) 55 { 56 for (int i=0;i<m;i++) 57 { 58 for (int j=0;j<n;j++) 59 { 60 cin>>maze[i][j]; 61 if (maze[i][j]=='r') 62 { 63 r.x=i;r.y=j;r.time=0; 64 } 65 } 66 } 67 mtime=9999; 68 dfs(r); 69 cout<<mtime<<endl; 70 71 } 72 return 0; 73 } 74 75 BFS+优先队列： 76 #include <iostream> 77 #include <stdio.h> 78 #include <queue> 79 80 /* 81 如何求最短的路径？？ 82 题意：X代表卫兵，a代表终点，r代表起始点，.代表路，#代表墙 83 84 路花费一秒，x花费两秒 85 86 问到达终点的最少时间 87 */ 88 using namespace std; 89 int m,n,xx,yy; 90 char maze[201][201]; 91 int go[4][2]= 92 { 93 -1,0, 94 0,1, 95 1,0, 96 0,-1 97 }; 98 struct node 99 { 100 int x,y,time; 101 /* bool operator< (const node &a)const 102 { 103 return time > a.time; 104 } 105 */ 106 }a,r; 107 bool operator < (node c,node d) 108 { 109 return c.time > d.time; 110 } 111 void bfs() 112 { 113 priority_queue<node> q; //默认大顶堆 114 q.push(r); 115 maze[r.x][r.y]='#'; 116 node t,newt; 117 while (!q.empty()) 118 { 119 t=q.top();q.pop(); 120 //--------------------!!!!!--------------------- 121 maze[t.x][t.y]='#'; 122 //至关重要的一句！目的是让走过的路不要再重复 123 //--------------------!!!!!--------------------- 124 for (int i=0;i<4;i++) 125 { 126 xx=t.x+go[i][0]; 127 yy=t.y+go[i][1]; 128 newt.x=xx; 129 newt.y=yy; 130 newt.time=t.time+1; 131 if (xx<0||yy<0||xx>=m||yy>=n||(maze[xx][yy]=='#')) continue; 132 if (maze[xx][yy]=='x') 133 { 134 newt.time++; 135 } 136 else 137 if (maze[xx][yy]=='a') 138 { 139 printf("%d ",newt.time); 140 return; 141 } 142 q.push(newt); 143 // maze[xx][yy]='#'; 这句话加在这里就会出错：不能把它放进去的时候就变路 144 } 145 } 146 printf("Poor ANGEL has to stay in the prison all his life. "); 147 148 } 149 150 int main() 151 { 152 153 while (scanf("%d%d",&m,&n)!=EOF) 154 { 155 for (int i=0;i<m;i++) 156 { 157 for (int j=0;j<n;j++) 158 { 159 cin>>maze[i][j]; 160 if (maze[i][j]=='r') 161 { 162 r.x=i;r.y=j;r.time=0; 163 } 164 } 165 } 166 bfs(); 167 168 } 169 return 0; 170 }
关于优先队列：
默认大顶堆，也就是队首为最大值。队列按照从大到小的顺序排列。如果需要从小到大排序：
两种方式：
- 结构体内运算符重载，记住括号右侧const必须要加不然会报错
- 结构体外运算符重载
- cmp结构体内运算符重载
- 友元函数
以结构体Time为例：
struct Time{ int start, end; };
使用优先队列时，如果需要对Time中的start从小到大排序，有两种方法：
priority_queue<Time> pq;
一.在结构体外重载结构体小于运算符：
bool operator <(const Time& a,const Time& b){ return a.start > b.start; } //这里以大于重载小于是因为默认情况下，优先队列是以大的作为队首，这样一反，就可以再默认情况下使得小的作为队首
二.直接在结构体中重载小于运算符：
struct Time{ int start, end; bool operator < (const Time& t)const{ return start > t.start; } };
// 第二个const的作用：（我已经百度了无数遍了），const函数，不能修改成员变量，括号内的const表示不能修改参数变量
三.直接在cmp中重载小于运算符：
struct cmp { bool operator <(Time t1, Time t2) { return t1.start > t2.start; } }; priority_queue<Time,vector<Time> , cmp > q; 调用形式
四.友元函数
在Time结构体中定义友元函数,省去了const的麻烦
friend bool operator< (Time f1, Time f2) { return f1.start < f2.start; }
什么是友元函数？
- 友元函数不属于类的成员函数，需要在类外实现函数主题，不需要加类限定
- 友元函数可以访问类的私有成员但必须通过类对象形参的方式使用
小顶堆：priority_queue<Time,vector<Time>,greater<Time>