Tree-shaped DP Water Quiz

BZOJ1131: [Poi2008]sta

Test instructions: Give a tree of n points, and find a point to which the root of the tree, the sum of the depths of all points is the maximum.

The puzzle: Record the depth of each point, and then derive the depth and level of the other points based on the depth of the root node and the layers.

I'm using BFS.

#include <stdio.h>#include<string.h>#include<iostream>#include<queue>using namespaceStd;typedefLong Longll;intN,cnt,ans;inthead[1000010],next[2000010],to[2000010],fa[1000010],s[1000010];ll f[1000010],g[1000010],siz[1000010];queue<int>Q;voidAddintAintb) {to[cnt]=b; NEXT[CNT]=Head[a]; Head[a]=cnt++;}intMain () {scanf ("%d",&N); memset (Head,-1,sizeof(head)); intI,a,b,u;  for(i=1; i<n;i++) {scanf ("%d%d",&a,&b);        Add (A, b);    Add (B,a); } Q.push (1);  while(!Q.empty ()) {u=Q.front ();        Q.pop (); s[++s[0]]=T; Siz[u]++;  for(i=head[u];i!=-1; i=Next[i]) {            if(to[i]!=Fa[u]) {Fa[to[i]]=u;            Q.push (To[i]); }        }    }     for(i=n;i>=2; i--) {siz[fa[s[i] ]+=Siz[s[i]]; F[fa[s[i] ]+=siz[s[i]]+f[s[i]];//F[i] Represents the sum of all sons of I to the depth of I} g[1]=f[ans=1];  for(i=2; i<=n;i++) {G[s[i]]=g[fa[s[i]]]+n-(siz[s[i]]<<1);//G[i] Represents the sum of all points to I, which can be launched by G[fa[i]]        if(g[ans]<g[s[i]]| | (g[ans]==g[s[i]]&&ans>S[i])) Ans=S[i]; } printf ("%d", ans); return 0;}

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Poj3107:godfather

Test instructions: Find a point in the tree so that when this point is removed, the maximum number of all remaining subtree nodes is minimized, to satisfy all the required nodes.

Solution: Find out the number of nodes per subtree size[x], the answer is min (size[y],n-size[x]), Y is the son of X.

#include <stdio.h>#include<string.h>#include<iostream>using namespacestd;Const intmaxn=50010;intN,cnt,minn;inthead[maxn],to[maxn<<1],next[maxn<<1],S[MAXN],F[MAXN],STA[MAXN];intReadin () {intret=0;CharGC;  while(gc<'0'|| Gc>'9') gc=GetChar ();  while(gc>='0'&&gc<='9') ret=ret*Ten+gc-'0', gc=GetChar (); returnret;}voidAddintAintb) {to[cnt]=b; NEXT[CNT]=Head[a]; Head[a]=cnt++;}voidDfsintXintFA) {S[x]=1;  for(inti=head[x];i!=-1; i=Next[i]) {        if(to[i]!=FA)            {DFS (TO[I],X); S[X]+=S[to[i]]; F[X]=Max (F[x],s[to[i]]); }} F[x]=max (f[x],n-s[x]); Minn=min (minn,f[x]);}intMain () {n=Readin (); memset (Head,-1,sizeof(head)); inti,a,b;  for(i=1; i<n;i++) {a=readin (), b=Readin ();    Add (A, B), add (B,a); } Minn=1<< -; DFS (1,0);  for(i=1; i<=n;i++)if(F[i]==minn) sta[++sta[0]]=i;  for(i=1; i<sta[0];i++) printf ("%d", Sta[i]); printf ("%d", sta[sta[0]]); return 0;}

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Poj1655:balancing Act

Test instructions: With POJ3107

#include <stdio.h>#include<iostream>#include<string.h>using namespacestd;Const intmaxn=20010;intn,minn,ans,cnt;ints[maxn],to[maxn<<1],next[maxn<<1],HEAD[MAXN];intReadin () {intret=0;CharGC;  while(gc<'0'|| Gc>'9') gc=GetChar ();  while(gc>='0'&&gc<='9') ret=ret*Ten+gc-'0', gc=GetChar (); returnret;}voidAddintAintb) {to[cnt]=b; NEXT[CNT]=Head[a]; Head[a]=cnt++;}voidDfsintXintFA) {    intt=0; S[X]=1;  for(inti=head[x];i!=-1; i=Next[i]) {        if(to[i]!=FA)            {DFS (TO[I],X); S[X]+=S[to[i]]; T=Max (T,s[to[i]]); }} t=max (t,n-s[x]); if(t<minn| |    (T==minn&&x<ans)) ans=x,minn=t;}voidWork () {n=Readin (); memset (Head,-1,sizeof(head)); Ans=cnt=0, minn=1<< -; inti,a,b;  for(i=1; i<n;i++) {a=readin (), b=Readin ();    Add (A, B), add (B,a); } DFS (1,0); printf ("%d%d\n", Ans,minn);}intMain () {intt=Readin ();  while(t--) work (); return 0;}

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bzoj1827:[usaco2010 Mar]gather Cow Big Rally

Test instructions: Similar to the previous question, except that different points have different cattle to live in, asking for the smallest total distance from all cows to the core

The puzzle: See http://www.cnblogs.com/CQzhangyu/p/6178852.html

Tree-shaped DP Water Quiz