Bzoj 2525 [poi2011]dynamite two points + tree-type greedy

Test Instructions:
n points, a tree, some points are key points, can be M-Dianran color.
Minimize the maximum distance from all key points to the nearest dyed point.
parsing:
Anyway, from this problem I have learned a kind of idea of doing the problem?
Grandpa said when the lecture: The general choice of some points of the same price is greedy, the price of different words are generally DP.
Think is also quite right, but is not aware of it?
Anyway, this exam I was direct D---!
Forget why D.
Obviously the maximum value is minimal. We need two minutes for this value.
Then we sweep the whole tree from the bottom up.
How many nodes are there in the state?
The first is that there are no key points within the subtree that are not overwritten, and the contribution of one node in the subtree can continue upward.
The second is that there are key points in the subtree that are not covered, and the contribution of the subtree without nodes can be continued upward.
The third is that there are no key points to be overwritten, and there is no point that can continue to contribute.
The fourth type is all there.
But we can prove that the fourth case exists, obviously can pick a point outside the subtree to cover the key is not covered, but this time the sub-tree in the selection of the point is not used, so this situation can be attributed to the third kind.
Then it is the process of greedy discussion.
The greedy thought is that you can dye it without dyeing it.
Obviously the first one we need to keep track of how much we can go up.
The second is the distance we need to record the farthest non-covered key points to reach the current root node

I think the way to do something is more important.
That's still the word.
the common choice of some points of the same price is greedy, the price of the different words are generally DP.
Perhaps this sentence will solve my future greed and DP trouble? 2333
Code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 300100using namespace STD;intN,m;intFlag[n];intHead[n];intStatus[n];intDocument[n];intCntintTotstructnode{intFrom,to,next;} edge[n<<1];voidInit () {memset(head,-1,sizeof(head)); Cnt=1;}voidEdgeadd (intFromintTo) {edge[cnt].from=from,edge[cnt].to=to;    Edge[cnt].next=head[from]; head[from]=cnt++;}//0-> Continue delivery//1-> has not been overwritten//2-> neither delivered nor not overwritten. voidDfsintNowintFaintDIS) {intuncovered=flag[now]-1;intnear=-1; for(inti=head[now];i!=-1; i=edge[i].next) {intto=edge[i].to;if(TO==FA)Continue;    DFS (To,now,dis); } for(inti=head[now];i!=-1; i=edge[i].next) {intto=edge[i].to;if(TO==FA)Continue;if(status[to]==0) Near=max (near,document[to]-1);Else if(status[to]==1) Uncovered=max (uncovered,document[to]+1); }if(near<uncovered) {if(Uncovered==dis) document[now]=dis,status[now]=0, tot++;Elsedocument[now]=uncovered,status[now]=1; }Else if(near!=-1) document[now]=near,status[now]=0;Elsestatus[now]=2, document[now]=0;}BOOLCheckintdis) {tot=0; Dfs1,0, dis);if(status[1]==1) tot++;returnTot<=m?1:0;}intMain () {init ();scanf("%d%d", &n,&m); for(intI=1; i<=n;i++)scanf("%d", &flag[i]); for(intI=1; i<n;i++) {intx, y;scanf("%d%d", &x,&y);        Edgeadd (x, y);    Edgeadd (Y,X); }intL=0, R=n,ans; while(L&LT;=R) {intMid= (l+r) >>1;if(Check (mid)) ans=mid,r=mid-1;ElseL=mid+1; }printf("%d\n", ans);}

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Bzoj 2525 [poi2011]dynamite binary + Tree greedy