本文主要是介绍Yaoge’s maximum profit HDU - 5052，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
http://acm.hdu.edu.cn/showproblem.php?pid=5052
和之前做得求链上连续段数很像 但是这道题难在要考虑方向问题 写起来很恶心
从u和v两边往上爬 在某一边每爬一小段(线段树上一个区间 需要区分方向) 就看这一小段上的利润最大值 再和两边已经走过部分的最大最小值做差取最优
#include <bits/stdc++.h>
using namespace std;
const int N=0x3f3f3f3f;struct node1
{int v;int next;
};struct node2
{int l;int r;int laz;int minn;int maxx;int val0;//leftint val1;//right
};node1 edge[100010];
node2 tree[200010];
int val[50010],first[50010],fa[50010],deep[50010],sum[50010],son[50010],top[50010],mp1[50010],mp2[50010];
int n,q,num;void addedge(int u,int v)
{edge[num].v=v;edge[num].next=first[u];first[u]=num++;
}void dfsI(int cur)
{int i,v;sum[cur]=1,son[cur]=-1;for(i=first[cur];i!=-1;i=edge[i].next){v=edge[i].v;if(v!=fa[cur]){fa[v]=cur,deep[v]=deep[cur]+1;dfsI(v);sum[cur]+=sum[v];if(son[cur]==-1||sum[son[cur]]<sum[v]){son[cur]=v;}}}
}void dfsII(int cur,int tp)
{int i,v;num++;top[cur]=tp,mp1[cur]=num,mp2[num]=cur;if(son[cur]==-1) return;dfsII(son[cur],tp);for(i=first[cur];i!=-1;i=edge[i].next){v=edge[i].v;if(v!=fa[cur]&&v!=son[cur]){dfsII(v,v);}}
}void pushup(int cur)
{tree[cur].minn=min(tree[2*cur].minn,tree[2*cur+1].minn);tree[cur].maxx=max(tree[2*cur].maxx,tree[2*cur+1].maxx);tree[cur].val0=max(max(0,tree[2*cur].maxx-tree[2*cur+1].minn),max(tree[2*cur].val0,tree[2*cur+1].val0));tree[cur].val1=max(max(0,tree[2*cur+1].maxx-tree[2*cur].minn),max(tree[2*cur].val1,tree[2*cur+1].val1));
}void pushdown(int cur)
{if(tree[cur].laz!=0){tree[2*cur].minn+=tree[cur].laz;tree[2*cur].maxx+=tree[cur].laz;tree[2*cur].laz+=tree[cur].laz;tree[2*cur+1].minn+=tree[cur].laz;tree[2*cur+1].maxx+=tree[cur].laz;tree[2*cur+1].laz+=tree[cur].laz;tree[cur].laz=0;}
}void build(int l,int r,int cur)
{int m;tree[cur].l=l;tree[cur].r=r;tree[cur].laz=0;if(l==r){tree[cur].minn=val[mp2[l]];tree[cur].maxx=val[mp2[l]];tree[cur].val0=0;tree[cur].val1=0;return;}m=(l+r)/2;build(l,m,2*cur);build(m+1,r,2*cur+1);pushup(cur);
}int queryII(int pl,int pr,int op,int &tminn,int &tmaxx,int cur)
{int res,lminn,lmaxx,rminn,rmaxx;if(pl<=tree[cur].l&&tree[cur].r<=pr){tminn=tree[cur].minn;tmaxx=tree[cur].maxx;if(op==0) return tree[cur].val0;else return tree[cur].val1;}pushdown(cur);if(pr<=tree[2*cur].r) return queryII(pl,pr,op,tminn,tmaxx,2*cur);else if(pl>=tree[2*cur+1].l) return queryII(pl,pr,op,tminn,tmaxx,2*cur+1);else{res=0;res=max(res,queryII(pl,pr,op,lminn,lmaxx,2*cur));res=max(res,queryII(pl,pr,op,rminn,rmaxx,2*cur+1));tminn=min(lminn,rminn);tmaxx=max(lmaxx,rmaxx);if(op==0) res=max(res,lmaxx-rminn);else res=max(res,rmaxx-lminn);return res;}
}int queryI(int u,int v)
{int res,minnu,maxxu,minnv,maxxv,minnt,maxxt;res=0,maxxu=0,minnu=N,maxxv=0,minnv=N;while(top[u]!=top[v]){if(deep[top[u]]>deep[top[v]]){res=max(res,queryII(mp1[top[u]],mp1[u],0,minnt,maxxt,1));res=max(res,maxxt-minnu);res=max(res,maxxv-minnt);minnu=min(minnu,minnt);maxxu=max(maxxu,maxxt);u=fa[top[u]];}else{res=max(res,queryII(mp1[top[v]],mp1[v],1,minnt,maxxt,1));res=max(res,maxxv-minnt);res=max(res,maxxt-minnu);minnv=min(minnv,minnt);maxxv=max(maxxv,maxxt);v=fa[top[v]];}}if(deep[u]>deep[v]){res=max(res,queryII(mp1[v],mp1[u],0,minnt,maxxt,1));res=max(res,maxxt-minnu);res=max(res,maxxv-minnt);}else{res=max(res,queryII(mp1[u],mp1[v],1,minnt,maxxt,1));res=max(res,maxxv-minnt);res=max(res,maxxt-minnu);}return res;
}void updateII(int pl,int pr,int val,int cur)
{if(pl<=tree[cur].l&&tree[cur].r<=pr){tree[cur].minn+=val;tree[cur].maxx+=val;tree[cur].laz+=val;return;}pushdown(cur);if(pl<=tree[2*cur].r) updateII(pl,pr,val,2*cur);if(pr>=tree[2*cur+1].l) updateII(pl,pr,val,2*cur+1);pushup(cur);
}void updateI(int u,int v,int w)
{while(top[u]!=top[v]){if(deep[top[u]]<deep[top[v]]) swap(u,v);updateII(mp1[top[u]],mp1[u],w,1);u=fa[top[u]];}if(deep[u]<deep[v]) swap(u,v);updateII(mp1[v],mp1[u],w,1);
}int main()
{int t,u,v,w,i;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&val[i]);}memset(first,-1,sizeof(first));num=0;for(i=1;i<=n-1;i++){scanf("%d%d",&u,&v);addedge(u,v);addedge(v,u);}fa[1]=-1,deep[1]=1;dfsI(1);num=0;dfsII(1,1);build(1,n,1);scanf("%d",&q);while(q--){scanf("%d%d%d",&u,&v,&w);printf("%d\n",queryI(u,v));updateI(u,v,w);}}return 0;
}
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