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BZOJ4813 CQOI2017小Q的棋盤(樹形dp)

bzoj online truct color 轉移 struct con div ons

  設f[i][j]為由i號點開始在子樹內走j步最多能經過多少格點,g[i][j]為由i號點開始在子樹內走j步且回到i最多能經過多少格點,轉移顯然。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 110
char getc(){char c=getchar();while
((c<A||c>Z)&&(c<a||c>z)&&(c<0||c>9)) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<0||c>9) {if (c==-) f=-1;c=getchar();} while (c>=0&&c<=9) x=(x<<1
)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,m,p[N],f[N][N],g[N][N],t; struct data{int to,nxt; }edge[N<<1]; void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;} void dfs(int k,int from) { f[k][0]=g[k][0]=1;for (int i=1;i<=m;i++) f[k][i]=g[k][i]=-n; for (int
i=p[k];i;i=edge[i].nxt) if (edge[i].to!=from) { dfs(edge[i].to,k); for (int x=m;x>=1;x--) for (int y=0;y+1<=x;y++) f[k][x]=max(f[k][x],max(g[k][x-y-1]+f[edge[i].to][y],(x-y-2>=0?f[k][x-y-2]+g[edge[i].to][y]:-n))); for (int x=m;x>=2;x--) for (int y=0;y+2<=x;y++) g[k][x]=max(g[k][x],g[k][x-y-2]+g[edge[i].to][y]); } } int main() { #ifndef ONLINE_JUDGE freopen("bzoj4813.in","r",stdin); freopen("bzoj4813.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif n=read(),m=read(); for (int i=1;i<n;i++) { int x=read()+1,y=read()+1; addedge(x,y),addedge(y,x); } dfs(1,1); for (int i=0;i<m;i++) f[1][m]=max(f[1][m],f[1][i]); cout<<f[1][m]; return 0; }

BZOJ4813 CQOI2017小Q的棋盤(樹形dp)