hdu5834 Magic boy Bi Luo with his excited tree(树形dp)
发布时间:2021-01-24 16:22:37 所属栏目:大数据 来源:网络整理
导读:副标题#e# Magic boy Bi Luo with his excited tree Time Limit: 8000/4000 MS (Java/Others)????Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 723????Accepted Submission(s): 192 Problem Description ? Bi Luo is a magic boy,he
? #pragma comment(linker,"/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <cmath> #include <algorithm> using namespace std; const double eps = 1e-6; const double pi = acos(-1.0); const int INF = 1e9; const int MOD = 1e9+7; #define ll long long #define CL(a,b) memset(a,b,sizeof(a)) #define lson (i<<1) #define rson ((i<<1)|1) #define N 100010 int gcd(int a,int b){return b?gcd(b,a%b):a;} struct node { int v,c,next; }e[N<<1]; int tot,head[N]; void add(int u,int v,int c) { e[tot].v = v; e[tot].c = c; e[tot].next = head[u]; head[u] = tot++; } int val[N]; int d_down[2][N],d_up[2][N]; ///dp_down[0/1][u]:u结点往其为根的子树走,并且不走回来/走回来,能得到的最大权值 ///dp_up[0/1][u]:u结点往其父亲向上走,并且不走回来/走回来,能得到的最大权值 void dfs1(int u,int fa) { d_down[0][u] = d_down[1][u] = val[u]; for(int i=head[u]; i!=-1; i=e[i].next) { int v = e[i].v; if(v == fa) continue; dfs1(v,u); if(d_down[0][v]-2*e[i].c>0) d_down[0][u] += d_down[0][v]-2*e[i].c; } int mx = 0; for(int i=head[u]; i!=-1; i=e[i].next) { int v = e[i].v; if(v == fa) continue; if(d_down[0][v]-2*e[i].c>0) mx = max(mx,(d_down[1][v]-e[i].c)-(d_down[0][v]-2*e[i].c)); else mx = max(mx,d_down[1][v]-e[i].c); } d_down[1][u] = d_down[0][u] + mx; } void dfs2(int u,int fa) { int mx1=0,mx2=0,tmp; for(int i=head[u]; i!=-1; i=e[i].next) { int v = e[i].v; if(v == fa) continue; if(d_down[0][v]-2*e[i].c>0) tmp=(d_down[1][v]-e[i].c)-(d_down[0][v]-2*e[i].c); else tmp=d_down[1][v]-e[i].c; if(mx1<tmp) mx2=mx1,mx1=tmp; else if(mx2<tmp) mx2=tmp; } for(int i=head[u]; i!=-1; i=e[i].next) { int v = e[i].v; if(v == fa) continue; int tmp2; if(d_down[0][v]-2*e[i].c>0) tmp2=d_down[0][u]-(d_down[0][v]-2*e[i].c); else tmp2=d_down[0][u]; int mx=max(d_up[0][u]-2*e[i].c,tmp2-2*e[i].c); mx = max(mx,d_up[0][u]+tmp2-2*e[i].c-val[u]); d_up[0][v] = val[v]+max(0,mx); if(d_down[0][v]-2*e[i].c>0) { if(mx1==(d_down[1][v]-e[i].c)-(d_down[0][v]-2*e[i].c)) tmp = d_down[1][u]-(d_down[1][v]-e[i].c)+mx2; else tmp = d_down[1][u]-(d_down[0][v]-2*e[i].c); } else if(d_down[1][v]-e[i].c>0) { if(mx1==d_down[1][v]-e[i].c) tmp = d_down[1][u]-(d_down[1][v]-e[i].c)+mx2; else tmp = d_down[1][u]; } else tmp = d_down[1][u]; mx = max(d_up[1][u]-e[i].c,tmp-e[i].c); mx = max(mx,max(d_up[0][u]+tmp-e[i].c-val[u],d_up[1][u]+tmp2-e[i].c-val[u])); d_up[1][v] = val[v]+max(0,mx); dfs2(v,u); } } int main() { int T; scanf("%d",&T); for(int cas=1; cas<=T; cas++) { tot = 0; CL(head,-1); int n; scanf("%d",&n); for(int i=1; i<=n; i++) scanf("%d",val+i); int a,c; for(int i=1; i<n; i++) { scanf("%d%d%d",&a,&b,&c); add(a,c); add(b,a,c); } dfs1(1,1); d_up[0][1] = d_up[1][1] = val[1]; dfs2(1,1); printf("Case #%d:n",cas); for(int i=1; i<=n; i++) { printf("%dn",max(d_up[1][i]+d_down[0][i],d_up[0][i]+d_down[1][i])-val[i]); } } return 0; } (编辑:PHP编程网 - 黄冈站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |