Maximum independent set of trees

Select a node as root, use D (i) to indicate the maximum independent set size of the subtree with I as the root node, and node I have only two decisions: Select or not. If I is selected, the son of I is not selectable, and if I is not selected, then the problem is converted to the D value of all sons who seek I. State transition equation:

where S (i) and GS (i) are the sons of I are set up with grandchildren.
Implementation method: When a D (i) is computed, it is used to update the parent node of I and the grandfather node, so that whenever the parent node can be recorded
poj2342 Anniversary Party

#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#include <string>#include <sstream>#include <vector>#include <set>#include <map>#include <queue>#include <deque>#include <stack>using namespace STD;Const intmaxn=6000+Ten; vector<int>NEXT[MAXN];intS[MAXN],GS[MAXN];intW[MAXN];intNintdpintCurintFA) { for(intI=0; I<next[cur].size (); i++) {intX=next[cur][i];if(X==FA)Continue;intD=DP (x,cur); S[cur]+=d;if(fa!=-1) Gs[fa]+=d; }returnMax (S[cur],gs[cur]+w[cur]);}intMain () { while(scanf("%d%d", &n,&w[1]), n| | w[1]){ for(intI=1; i<=n;i++) next[i].clear ();intx, y; for(intI=2; i<=n;i++)scanf("%d", &w[i]); for(intI=1; i<=n-1; i++) {scanf("%d%d", &x,&y);            Next[x].push_back (y);        Next[y].push_back (x); }memset(S,0,sizeof(s));memset(GS,0,sizeof(GS));printf("%d\n", DP (1,-1)); }return 0;}

poj3342 Party at Hali-bula
Map names to numbers faster.
It's a bit of a hassle to check if the answer is unique.

#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#include <string>#include <sstream>#include <vector>#include <set>#include <map>#include <queue>#include <deque>#include <stack>using namespace STD;Const intmaxn= $+Ten; Map<string,int>Id vector<int>NEXT[MAXN];intS[MAXN],GS[MAXN];intD[MAXN];intNvoiddpintCurintFA) { for(intI=0; I<next[cur].size (); i++) {intX=next[cur][i];if(X==FA)Continue;        DP (X,CUR); S[CUR]+=D[X];if(fa!=-1) gs[fa]+=d[x]; } D[cur]=max (s[cur],gs[cur]+1);}BOOLDfsintCurintFA) {if(next[cur].size () = =0)return true;if(d[cur]==s[cur]&&d[cur]==gs[cur]+1)return false; for(intI=0; I<next[cur].size (); i++) {intX=next[cur][i];if(X==FA)Continue;if(s[cur]>gs[cur]+1){//If the child node is taken            if(Dfs (x,cur) = =false)return false; }Else{//If you are taking a grandson knot.             for(intj=0; J<next[x].size (); j + +) {intY=NEXT[X][J];if(y==cur)Continue;if(Dfs (y,x) = =false)return false; }        }    }return true;}intMain () {stringPeox,peoy; while(scanf("%d", &n), N) {id.clear (); for(intI=1; i<=n;i++) next[i].clear ();intCnt=0;Cin>>peox; for(intI=1; i<=n-1; i++) {Cin>>peox>>peoy;if(Id.count (peox) = =0) id[peox]=++cnt;if(Id.count (peoy) = =0) id[peoy]=++cnt;intX=id[peox],y=id[peoy];            Next[x].push_back (y);        Next[y].push_back (x); }memset(d,0,sizeof(d));memset(S,0,sizeof(s));memset(GS,0,sizeof(GS)); dp1,-1);printf("%d", d[1]);puts(Dfs (1,-1)?"Yes":"No"); }return 0;}

Maximum independent set of trees