标签：enter using 节点 stdin 预处理 turn 题解 sizeof ast
Time Limit: 20 Sec
Memory Limit: 256 MB
http://acm.hdu.edu.cn/showproblem.php?pid=4123
Input
There are several test cases.
The first line of each test case contains two integers N and M. N is the number of houses, M is the number of queries.
The following N-1 lines, each contains three integers, x, y
and z, indicating that there is a road of length z connecting house x
and house y.
The following M lines are the queries. Each line contains an
integer Q, asking that at most how many people can take part in Bob’s
race according to the above mentioned rules and under the condition that
the“race difference”is no more than Q.
The input ends with N = 0 and M = 0.
(N<=50000 M<=500 1<=x,y<=N 0<=z<=5000 Q<=10000000)
Output
For each test case, you should output the answer in a line for each query.
Sample Input
Sample Output
1
3
3
3
5
题意
一个城镇有N个住户，N-1条路连接两个住户，保证N个住户联通，M次询问，给定N条边的信息，包括连
接的住户序号以及路的长度。然后是M次询问，每次询问Q，要求找到最长的连续序号(我一开始看成了树上连续一段），使得Max(dis[i]) - Min(dis[i]) ≤
Q(l≤i≤r)，输出最大的r-l+1。dis[i]为从第i个住户出发，不重复走过路能移动的最远距离。
题解：
１．首先需要知道一个 结论：树上每个节点理他最远的点肯定是数的直径的两个端点的其中一个。
２．如果你不会求树的直径以及两个端点，没关系，其实不难，任意找一个点dfs找到离它最远的点root1，root1一定是其中一个端点，然后再以root1为根dfs一遍求出理root1最远的点root2。这样就找到了两个端点。
３．求出每个点到最远点的距离，我这里是分别以root1,root2 dfs，求出每个点到root1和root2的距离，取个max即可。
４．题目转化成了，求最长的区间(l,r),满足max{l,r}-min{l,r}<=q。
５．其实可以用尺取法来做，因为假如(l,r)区间合法，那么(l+1,r)一定合法，那么枚举左端点，右端点只能单增，时间复杂度是O(n)的。
６．最后要搞个ST表求区间最大最小值。
我一开始用优先队列做的，然后T了，真是花样作死，非要加个log。
衷心提示:k=(int)(log(y-x+1)/2)会超时，这个需要预处理，亲测一个1.5s，另一个5.2秒。
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define N 50050 5 int n,m,val[N],dp1[N],dp2[N],flag[N]; 6 int tot,last[N]; 7 struct Edge{int from,to,val,s;}edges[N<<1]; 8 struct Node 9 { 10 int id,val; 11 bool operator <(const Node&b)const 12 {return val<b.val;} 13 }; 14 template<typename T>void read(T&x) 15 { 16 ll k=0; char c=getchar(); 17 x=0; 18 while(!isdigit(c)&&c!=EOF)k^=c==‘-‘,c=getchar(); 19 if (c==EOF)exit(0); 20 while(isdigit(c))x=x*10+c-‘0‘,c=getchar(); 21 x=k?-x:x; 22 } 23 void read_char(char &c) 24 {while(!isalpha(c=getchar())&&c!=EOF);} 25 void AddEdge (int x,int y,int z) 26 { 27 edges[++tot]=Edge{x,y,z,last[x]}; 28 last[x]=tot; 29 } 30 void dfs1(int x,int pre,int &mx,int &root) 31 { 32 mx=0; root=x; 33 int tpmx,tproot; 34 for(int i=last[x];i;i=edges[i].s) 35 { 36 Edge &e=edges[i]; 37 if (e.to==pre)continue; 38 dfs1(e.to,x,tpmx,tproot); 39 if (tpmx+e.val>mx) 40 { 41 mx=tpmx+e.val; 42 root=tproot; 43 } 44 } 45 } 46 void dfs2(int x,int pre,int *dp) 47 { 48 for(int i=last[x];i;i=edges[i].s) 49 { 50 Edge &e=edges[i]; 51 if (e.to==pre)continue; 52 dp[e.to]=dp[x]+e.val; 53 dfs2(e.to,x,dp); 54 } 55 } 56 57 void init() 58 { 59 read(n); read(m); 60 if (n==0&&m==0)exit(0); 61 for(int i=1;i<=n-1;i++) 62 { 63 int x,y,z; 64 read(x); read(y); read(z); 65 AddEdge(x,y,z); 66 AddEdge(y,x,z); 67 } 68 int maxn=0,root1,root2; 69 dfs1(1,0,maxn,root1); 70 dfs1(root1,0,maxn,root2); 71 dfs2(root1,0,dp1); 72 dfs2(root2,0,dp2); 73 for(int i=1;i<=n;i++) val[i]=max(dp1[i],dp2[i]); 74 } 75 void solve() 76 { 77 int r=0,ans=0,q; 78 read(q); 79 priority_queue<Node>Qmx,Qmi; 80 memset(flag,0,sizeof(flag)); 81 for(int i=1;i<=n;i++) 82 { 83 while(r+1<=n&&(Qmx.empty()|| 84 (fabs(val[r+1]-Qmx.top().val) 85 <=q&&fabs(val[r+1]+Qmi.top().val)<=q))) 86 { 87 Qmx.push(Node{r+1,val[r+1]}); 88 Qmi.push(Node{r+1,-val[r+1]}); 89 r++; 90 } 91 ans=max(ans,r-i+1); 92 flag[i]=1; 93 while(!Qmx.empty()&&flag[Qmx.top().id])Qmx.pop(); 94 while(!Qmi.empty()&&flag[Qmi.top().id])Qmi.pop(); 95 } 96 printf("%d\n",ans); 97 } 98 void clear() 99 { 100 tot=0; 101 memset(last,0,sizeof(last)); 102 memset(dp1,0,sizeof(dp1)); 103 memset(dp2,0,sizeof(dp2)); 104 memset(flag,0,sizeof(flag)); 105 } 106 int main() 107 { 108 #ifndef ONLINE_JUDGE 109 freopen("aa.in","r",stdin); 110 #endif 111 while(1) 112 { 113 clear(); 114 init(); 115 for(int i=1;i<=m;i++) solve(); 116 } 117 }
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define N 50050 5 int n,m,dp1[N],dp2[N],mx[N][20],mi[N][20],mm[N]; 6 int tot,last[N]; 7 struct Edge{int from,to,val,s;}edges[N<<1]; 8 template<typename T>void read(T&x) 9 { 10 ll k=0; char c=getchar(); 11 x=0; 12 while(!isdigit(c)&&c!=EOF)k^=c==‘-‘,c=getchar(); 13 if (c==EOF)exit(0); 14 while(isdigit(c))x=x*10+c-‘0‘,c=getchar(); 15 x=k?-x:x; 16 } 17 void read_char(char &c) 18 {while(!isalpha(c=getchar())&&c!=EOF);} 19 void AddEdge (int x,int y,int z) 20 { 21 edges[++tot]=Edge{x,y,z,last[x]}; 22 last[x]=tot; 23 } 24 void dfs1(int x,int pre,int &mx,int &root) 25 { 26 mx=0; root=x; 27 int tpmx,tproot; 28 for(int i=last[x];i;i=edges[i].s) 29 { 30 Edge &e=edges[i]; 31 if (e.to==pre)continue; 32 dfs1(e.to,x,tpmx,tproot); 33 if (tpmx+e.val>mx) 34 { 35 mx=tpmx+e.val; 36 root=tproot; 37 } 38 } 39 } 40 void dfs2(int x,int pre,int *dp) 41 { 42 for(int i=last[x];i;i=edges[i].s) 43 { 44 Edge &e=edges[i]; 45 if (e.to==pre)continue; 46 dp[e.to]=dp[x]+e.val; 47 dfs2(e.to,x,dp); 48 } 49 } 50 51 void init() 52 { 53 read(n); read(m); 54 if (n==0&&m==0)exit(0); 55 for(int i=1;i<=n-1;i++) 56 { 57 int x,y,z; 58 read(x); read(y); read(z); 59 AddEdge(x,y,z); 60 AddEdge(y,x,z); 61 } 62 int maxn=0,root1,root2; 63 dfs1(1,0,maxn,root1); 64 dfs1(root1,0,maxn,root2); 65 dfs2(root1,0,dp1); 66 dfs2(root2,0,dp2); 67 for(int i=1;i<=n;i++) mx[i][0]=mi[i][0]=max(dp1[i],dp2[i]); 68 } 69 void init_ST() 70 { 71 mm[0]=-1; 72 for(int i=1;i<=n;i++)mm[i]=(i&(i-1))==0?mm[i-1]+1:mm[i-1]; 73 for(int i=1;i<=20;i++) 74 for(int j=1;j+(1<<i)-1<=n;j++) 75 { 76 mx[j][i]=max(mx[j][i-1],mx[j+(1<<(i-1))][i-1]); 77 mi[j][i]=min(mi[j][i-1],mi[j+(1<<(i-1))][i-1]); 78 } 79 } 80 int rmq(int x,int y) 81 { 82 int k=mm[y-x+1]; 83 int ans=max(mx[x][k],mx[y-(1<<k)+1][k]); 84 ans-=min(mi[x][k],mi[y-(1<<k)+1][k]); 85 return ans; 86 } 87 void solve() 88 { 89 int r=0,ans=0,q; 90 read(q); 91 for(int i=1;i<=n;i++) 92 { 93 while(r+1<=n&&rmq(i,r+1)<=q)r++; 94 ans=max(ans,r-i+1); 95 } 96 printf("%d\n",ans); 97 } 98 void clear() 99 { 100 tot=0; 101 memset(last,0,sizeof(last)); 102 memset(dp1,0,sizeof(dp1)); 103 memset(dp2,0,sizeof(dp2)); 104 } 105 int main() 106 { 107 #ifndef ONLINE_JUDGE 108 freopen("aa.in","r",stdin); 109 #endif 110 clock_t now=clock(); 111 while(1) 112 { 113 clear(); 114 init(); 115 init_ST(); 116 for(int i=1;i<=m;i++) solve(); 117 } 118 119 }
标签：enter using 节点 stdin 预处理 turn 题解 sizeof ast
原文地址：https://www.cnblogs.com/mmmqqdd/p/10827457.html
