AtCoder Grand Contest 083 E - Bichrome Tree
阿新 • • 發佈:2018-12-05
題目傳送門:https://arc083.contest.atcoder.jp/tasks/arc083_c
題目大意:
給定一棵樹,你可以給這些點任意黑白染色,並且賦上權值,現給定一個序列\(X_i\),滿足對於每一個點\(i\),整棵子樹內所有和\(i\)顏色相同的點的權值和為\(X_i\),問是否可能
首先因為權值大小任意,所以\(v\)的子樹內權值和只要不超過\(X_v\)就好,那麼對於一個點\(v\)假定其為黑色,那麼子樹中黑色總和為\(X_v\),白色總和就要儘量小,定義為\(f_v\)
那麼選定\(v\)為黑色後,假定子樹內黑色總和為\(B\),白色總和為\(W\),那麼對於每個子節點\(v\)
\(B+=X_u,W+=f_u\)或\(B+=f_u,W+=X_u\)
/*program from Wolfycz*/ #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define inf 0x3f3f3f3f #define min(x,y) (x<y?x:y) #define max(x,y) (x>y?x:y) using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } inline int frd(){ int x=0,f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<3)+(x<<1)+ch-'0'; return x*f; } inline int read(){ int x=0,f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int N=1e3,M=5e3; int pre[N+10],now[N+10],child[N+10],tot; int f[N+10],g[2][M+10],v[N+10]; void join(int x,int y){pre[++tot]=now[x],now[x]=tot,child[tot]=y;} void dfs(int x){ for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]) dfs(son); memset(g[0],63,sizeof(g[0])); int T=0; g[T][0]=0; for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){ T^=1; memset(g[T],63,sizeof(g[T])); for (int i=0;i<=v[x];i++){ if (i>=v[son]) g[T][i]=min(g[T][i],g[T^1][i-v[son]]+f[son]); if (i>=f[son]) g[T][i]=min(g[T][i],g[T^1][i-f[son]]+v[son]); } } for (int i=0;i<=v[x];i++) f[x]=min(f[x],g[T][i]); } int main(){ int n=read(); for (int i=2;i<=n;i++) join(read(),i); for (int i=1;i<=n;i++) v[i]=read(); memset(f,63,sizeof(f)); dfs(1); printf(f[1]<inf?"POSSIBLE\n":"IMPOSSIBLE\n"); return 0; }