Codeforces 767C 树形dp
来源:互联网 发布:淘宝网电子测平仪 编辑:程序博客网 时间:2024/05/03 09:45
Once at New Year Dima had a dream in which he was presented a fairy garland. A garland is a set of lamps, some pairs of which are connected by wires. Dima remembered that each two lamps in the garland were connected directly or indirectly via some wires. Furthermore, the number of wires was exactly one less than the number of lamps.
There was something unusual about the garland. Each lamp had its own brightness which depended on the temperature of the lamp. Temperatures could be positive, negative or zero. Dima has two friends, so he decided to share the garland with them. He wants to cut two different wires so that the garland breaks up into three parts. Each part of the garland should shine equally, i. e. the sums of lamps' temperatures should be equal in each of the parts. Of course, each of the parts should be non-empty, i. e. each part should contain at least one lamp.
Help Dima to find a suitable way to cut the garland, or determine that this is impossible.
While examining the garland, Dima lifted it up holding by one of the lamps. Thus, each of the lamps, except the one he is holding by, is now hanging on some wire. So, you should print two lamp ids as the answer which denote that Dima should cut the wires these lamps are hanging on. Of course, the lamp Dima is holding the garland by can't be included in the answer.
The first line contains single integer n (3 ≤ n ≤ 106) — the number of lamps in the garland.
Then n lines follow. The i-th of them contain the information about the i-th lamp: the number lamp ai, it is hanging on (and 0, if is there is no such lamp), and its temperature ti ( - 100 ≤ ti ≤ 100). The lamps are numbered from 1 to n.
If there is no solution, print -1.
Otherwise print two integers — the indexes of the lamps which mean Dima should cut the wires they are hanging on. If there are multiple answers, print any of them.
62 40 54 22 11 14 2
1 4
62 40 64 22 11 14 2
-1
The garland and cuts scheme for the first example:
题意:给出一棵树,问能不能切去两条边,分成的三颗子树的点权和相同。
题解:首先算出sum,如果sum不能被3整除,肯定不行。
然后sum/3
dp[i]代表i点和i点的子树的权值和,所以从1结点开始dfs,dp记录,如果当前点的子树和自身的和为sum/3就删掉这颗子树。
如果在下的代码细节看不明白可以留言
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int head[3000005],cnt=0,n,num[3000005],dp[3000005],vis[3000005],ans[3],d,sum;struct node{int to,nex;}edge[6000005];void add(int u,int v){edge[cnt].to=v;edge[cnt].nex=head[u];head[u]=cnt++;}int flag;void travel(int s,int e,int fa){for(int i=head[s];~i;i=edge[i].nex){int v=edge[i].to;if(v==fa)continue;if(v==e){ans[++d]=i/2+1;flag=1;return ;}else{travel(v,e,s);if(flag)return ;}}}void dfs(int t){int i;for(i=head[t];~i;i=edge[i].nex){int v=edge[i].to;if(!vis[v]){vis[v]=1;dfs(v);dp[t]+=dp[v];if(d==2)return ;}}if(dp[t]==sum){flag=0;travel(1,t,1);dp[t]=0;}}int main(){int i,j,x,y;scanf("%d",&n);memset(head,-1,sizeof(head));for(i=1;i<=n;i++){scanf("%d%d",&x,&y);if(x!=0){add(i,x);add(x,i);}else cnt+=2;num[i]=dp[i]=y;sum+=y;}if(sum%3){printf("-1\n");return 0;}sum/=3;vis[1]=1;dfs(1);if(d>=2)printf("%d %d\n",ans[1],ans[2]);else printf("-1\n");return 0;}
- Codeforces 767C 树形dp
- 【CodeForces 767C】Garland (树形DP)
- Codeforces 77C 树形DP
- Codeforces 708C 树形DP
- Codeforces 77C 树形dp + 贪心
- Codeforces 708C Centroids(树形dp)
- codeforces contest 855 problem C(树形DP)
- CodeForces 238C World Eater Brothers 树形dp
- 【树形dp】Codeforces 238C World Eater Brothers
- Codeforces Round #148 (Div. 1)C 树形dp
- Codeforces 280C Game on Tree 树形期望dp
- Codeforces 455C Civilization(并查集+树形dp)
- 【Codeforces 736C】 Ostap and Tree【树形DP】
- Codeforces 369C Valera And Elections 树形DP
- Codeforces 815C. Karen and Supermarket 【树形DP】
- Codeforces 855 C. Helga Hufflepuff’s Cup (树形dp)
- CodeForces 855C Helga Hufflepuff's Cup(树形dp)
- codeforces 855C Helga Hufflepuff's Cup (树形dp)
- Android应用组件之ContentProvider
- WebApp方式集成5+ sdk
- ubuntu16.04 + GTX1060 + Cuda 8.0 + Zed SDK 环境搭建
- HTTP Header 详解
- java之JDK下载安装以及设置环境变量
- Codeforces 767C 树形dp
- HTML5学习
- 多线程基础(二)
- Linux date 时间戳和时间互转
- 【easyui学习笔记】4.动态添加标签页
- Codeforces Round #397 F Souvenirs(线段树)
- map遍历
- thinkphp中U方法按路由规则生成url
- Html发展史