D Tree Requests dfs+二分 D Pig and Palindromes -dp
Roman planted a tree consisting of n vertices. Each vertex contains a lowercase English letter. Vertex 1 is the root of the tree, each of then - 1 remaining vertices has a parent in the tree. Vertex is connected with its parent by an edge. The parent of vertex i is vertex pi, the parent index is always less than the index of the vertex (i.e., pi < i).
The depth of the vertex is the number of nodes on the path from the root to v along the edges. In particular, the depth of the root is equal to 1.
We say that vertex u is in the subtree of vertex v, if we can get from u to v, moving from the vertex to the parent. In particular, vertex v is in its subtree.
Roma gives you m queries, the i-th of which consists of two numbers vi, hi. Let's consider the vertices in the subtree vi located at depthhi. Determine whether you can use the letters written at these vertices to make a string that is a palindrome. The letters that are written in the vertexes, can be rearranged in any order to make a palindrome, but all letters should be used.
The first line contains two integers n, m (1 ≤ n, m ≤ 500 000) — the number of nodes in the tree and queries, respectively.
The following line contains n - 1 integers p2, p3, ..., pn — the parents of vertices from the second to the n-th (1 ≤ pi < i).
The next line contains n lowercase English letters, the i-th of these letters is written on vertex i.
Next m lines describe the queries, the i-th line contains two numbers vi, hi (1 ≤ vi, hi ≤ n) — the vertex and the depth that appear in the i-th query.
Print m lines. In the i-th line print "Yes" (without the quotes), if in the i-th query you can make a palindrome from the letters written on the vertices, otherwise print "No" (without the quotes).
题解：把相同深度的点存入一个数组中，每个点都是当前这个点的在树中查找的时间 在询问的时候就按照深度在相应的 数组中找出 时间区间在某个区间内的点就ok了
#include <iostream> #include <algorithm> #include <string.h> #include <vector> #include <cstdio> using namespace std; #define mk make_pair #define pub push_back const int maxn=500005; int num[maxn],depth[maxn],childmaxdepth[maxn]; int in[maxn],out[maxn],tim; vector<int>G[maxn]; char str[maxn]; vector< pair<int,int> >D[maxn]; void dfs(int cur,int de) { int siz=G[cur].size(); in[cur]=++tim; depth[cur]=childmaxdepth[cur]=de; if(D[de].size()<1){ D[de].pub(mk(0,0)); } int d=str[cur-1]-'a'; D[de].pub(mk(in[cur],D[de].back().second^(1<<d))); for(int i=0; i<siz; i++) { int to=G[cur][i]; dfs(to,de+1); childmaxdepth[cur]=max(childmaxdepth[cur],childmaxdepth[to]); } out[cur]=++tim; } int main() { int n,m; while(scanf("%d%d",&n,&m)==2) { tim=0; for(int i=2; i <= n; i++) { int d; scanf("%d",&d); G[d].push_back(i); } scanf("%s",str); dfs(1,1); for(int i=0; i<m; i++) { int val,c; scanf("%d%d",&val,&c); if(childmaxdepth[val]<c||c<=depth[val]) { puts("Yes");continue; } int L=lower_bound(D[c].begin(),D[c].end(), mk(in[val],-1))-D[c].begin(); int R=lower_bound(D[c].begin(),D[c].end(), mk(out[val],-1))-D[c].begin(); int t=D[c][L-1].second^D[c][R-1].second; int ok=t-(t&(-t)); if(ok)puts("No"); else puts("Yes"); } } return 0; }
E
给了一个矩阵 每个格子中有一个字符 然后要求从11 走到 nm 点 是一个回文串 只能从 向右或者向下走
我们枚举步数，然后 dp[x1][x2]可以得到 现在的点 (x1,y1), (x2,y2)，然后我们再次使用利用滚动数组可以得到想要的 从之前的那些点得到
//向左向右 add(dp[cur][x1][x2],dp[cur^1][x1][x2]);
//向左向上 add(dp[cur][x1][x2],dp[cur^1][x1][x2+1]);
//向下向右 add(dp[cur][x1][x2],dp[cur^1][x1-1][x2]);
//向下向上 add(dp[cur][x1][x2],dp[cur^1][x1-1][x2+1]);
#include <iostream> #include <cstdio> #include <algorithm> #include <string.h> using namespace std; const int maxn=505; long long dp[2][maxn][maxn]; char str[maxn][maxn]; const int mod=1000000007; int main() { int n,m; scanf("%d%d",&n,&m); for(int i=0; i<n; i++) scanf("%s",str[i]); if(n+m<=3){ printf("%d\n",str[0][0]==str[n-1][m-1]); return 0; } if(str[0][0]!=str[n-1][m-1]){ printf("%d\n",0);return 0; } for(int i=0; i<n; i++) for(int j=0; j<m; j++)dp[0][i][j]=0; dp[0][0][n-1]=1; int cur=0; for(int step=1; step<(n+m)/2;step++) { cur^=1; memset(dp[cur],0,sizeof(dp[cur])); for(int i=0; i < n&&i<=step; i++) for(int j=n-1; j>=0&&j>=i&&n-1-j<=step;j--) { if( ( step - i )> ( m - 1 - ( step-( n - 1 - j ) ) ) )continue; int x1=i,x2=j,y1=step-i,y2=m-1-(step-(n-1-j)); if(str[x1][y1]!=str[x2][y2])continue; dp[cur][x1][x2]=dp[cur^1][x1][x2]; dp[cur][x1][x2]=( dp[cur][x1][x2] + dp[cur^1][x1][x2+1] )%mod; if(x1>0) { dp[cur][x1][x2]=( dp[cur][x1][x2] + dp[cur^1][x1-1][x2] )%mod; dp[cur][x1][x2]=( dp[cur][x1][x2] + dp[cur^1][x1-1][x2+1] )%mod; } } } long long ans=0; for(int i=0; i<n; i++) ans=(ans+dp[cur][i][i])%mod; if( (n+m)%2){ for(int i=0; i<n-1; i++) ans=(ans+dp[cur][i][i+1])%mod; } printf("%I64d\n",ans); return 0; }
