hdu 1520 没有上司的晚会
树形dp入门题
传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1520
Attention：
hdu上每个输入都有多组数据
可能有既没有上司也没有下属的孤点
1 #include<cstdio> 2 #include<cstring> 3 int n,i,j,f[6010],fen[6010],l,k,ans=0,top=0; 4 int a[6010],b[6010],ok[6010]; 5 struct node{ 6 int v; 7 node *next; 8 }; 9 node *g[6010],*p; 10 void add(int u,int v) 11 { 12 p=new(node); 13 p->v=v; 14 p->next=g[u]; 15 g[u]=p; 16 } 17 int dui[6010],h,t; 18 int max(int a,int b) 19 { 20 if (a>b) return a;else return b; 21 } 22 int main() 23 { 24 while (scanf("%d",&n)!=EOF) 25 { 26 ans=0; 27 memset(a,0,sizeof(a)); 28 memset(b,0,sizeof(b)); 29 memset(ok,0,sizeof(ok)); 30 memset(f,0,sizeof(f)); 31 for (i=1;i<=n;i++) g[i]=NULL; 32 for (i=1;i<=n;i++) scanf("%d",&fen[i]); 33 if (n==1) {printf("%d\n",fen[1]); return 0;} 34 while (scanf("%d%d",&l,&k)) 35 { 36 if (l+k==0) break; 37 if (ok[k]==0) ok[k]=3; 38 if (ok[k]==1) ok[k]=2; 39 if (ok[l]==0) ok[l]=1; 40 if (ok[l]==3) ok[l]=2; 41 f[l]=k; add(k,l); 42 } 43 for (i=1;i<=n;i++) 44 { 45 if (ok[i]==0) ans+=fen[i]; 46 if (ok[i]==1) {a[i]=fen[i]; b[i]=0;} 47 if (ok[i]==2) a[i]=fen[i]; 48 if (ok[i]==3) {a[i]=fen[i]; top=i;} 49 } 50 51 dui[1]=top; h=0; t=1; 52 for (;h!=t;) 53 { 54 h++; 55 p=g[dui[h]]; 56 while (p!=NULL) 57 { 58 dui[++t]=p->v; 59 p=p->next; 60 } 61 } 62 63 for (i=t;i>1;i--) 64 { 65 int tmp=dui[i]; 66 a[f[tmp]]+=b[tmp]; //a[i]:i来时子树i的氛围值最大值 67 b[f[tmp]]+=max(a[tmp],b[tmp]); //b[i]:i不来时子树i的氛围值最大值 68 } 69 printf("%d\n",ans+max(a[top],b[top])); 70 } 71 }
