[POJ1155] [Croatia2002FinalExam] TELE [树形dp][分组背包]

[ L i n k \frak{Link} Link]

*本题 s t d \frak{std} std复杂度 Θ ( n 3 ) \frak{\Theta(n^3)} Θ(n3)卡过

解法如题。简单树形dp。

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cstdlib>#include<cmath>#include<ctime>#include<queue>#include<cctype>using namespace std;#define add_edge(a,b,c) nxt[++tot]=head[a],head[a]=tot,to[tot]=b,val[tot]=cint N,M,tot=0,ans=0;int head[3005]={},nxt[6005]={},to[6005]={},val[6005]={},F[3005][3005]={},siz[3005]={};void solve(int x){if(!head[x]){siz[x]=1;return;}for(int i=head[x];i;i=nxt[i]){solve(to[i]);siz[x]+=siz[to[i]];for(int j=siz[x];j>=1;--j){for(int k=j;k>=0;--k)//这层循环正反无所谓 {F[x][j]=max(F[x][j],F[x][k]+F[to[i]][j-k]-val[i]);}}}}int main(){scanf("%d%d",&N,&M);for(int i=0;i<=N;++i)for(int j=1;j<=M;++j)F[i][j]=-1000000000;for(int K,i=1;i<=N-M;++i){scanf("%d",&K);for(int A,C,j=1;j<=K;++j){scanf("%d%d",&A,&C);add_edge(i,A,C);}}for(int i=N-M+1;i<=N;++i)scanf("%d",&F[i][1]);solve(1);for(int i=siz[1];i;--i){if(F[1][i]>=0){printf("%d",i);return 0;}}return 0;}