树的最大独立集+判断唯一性
Party at Hali-Bula
Time Limit:3000MSMemory Limit:Unknown64bit IO Format:%lld & %llu
Submit
Description
Dear Contestant,I’m going to have a party at my villa at Hali-Bula to celebrate my retirement from BCM. I wish I could invite all my co-workers, but imagine how an employee can enjoy a party when he finds his boss among the guests! So, I decide not to invite both an employee and his/her boss. The organizational hierarchy at BCM is such that nobody has more than one boss, and there is one and only one employee with no boss at all (the Big Boss)! Can I ask you to please write a program to determine the maximum number of guests so that no employee is invited when his/her boss is invited too? I’ve attached the list of employees and the organizational hierarchy of BCM.
Best,-Brian Bennett
P.S. I would be very grateful if your program can indicate whether the list of people is uniquely determined if I choose to invite the maximum number of guests with that condition.Input
The input consists of multiple test cases. Each test case is started with a line containing an integern(1
n
200), the number of BCM employees. The next line contains the name of the Big Boss only. Each of the followingn- 1lines contains the name of an employee together with the name of his/her boss. All names are strings of at least one and at most 100 letters and are separated by blanks. The last line of each test case contains a single0.
Output
For each test case, write a single line containing a number indicating the maximum number of guests that can be invited according to the required condition, and a wordYesorNo, depending on whether the list of guests is unique in that case.
Sample Input6JasonJack JasonJoe JackJill JasonJohn JackJim Jill2MingCho Ming0Sample Output4 Yes1 No
Source
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: More Advanced Topics :: More Advanced DP Techniques ::DP level 4Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 9. Dynamic Programming ::ExamplesRoot :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: More Advanced Topics :: More Advanced Dynamic Programming ::The More Challenging Ones
Submit
/* ***********************************************Author:CKbossCreated Time :2015年02月15日 星期日 19时06分26秒File Name:UVA1220.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;const int maxn=300;int n,cnt;string name1,name2;vector<int> vi[maxn];map<string,int> msi;bool vised[maxn][2];struct Node{int val,unique;}dp[maxn][2];int getID(string name){if(msi[name]!=0) return msi[name];return msi[name]=cnt++;}void init(){msi.clear();cnt=1;for(int i=0;i<=n+10;i++) vi[i].clear();memset(vised,0,sizeof(vised));}Node dfs(int now,int kind){if(vised[now][kind]) return dp[now][kind];int sz=vi[now].size();/// 选择now时 dp[now][1]if(kind==1){if(sz==0){dp[now][1].val=1;dp[now][1].unique=1;vised[now][1]=true;return dp[now][1];}bool flag=true;int sum=1;for(int i=0;i<sz;i++) {int v=vi[now][i];Node nd = dfs(v,0);if(nd.unique==false) flag=false;sum+=nd.val;}dp[now][1].val=sum;dp[now][1].unique=flag;vised[now][1]=true;return dp[now][1];}/// 不选择now时if(kind==0){if(sz==0){dp[now][0].val=0;dp[now][0].unique=1;vised[now][0]=true;return dp[now][0];}bool flag=true;int sum=0;for(int i=0;i<sz;i++){int v=vi[now][i];Node nd0=dfs(v,0);Node nd1=dfs(v,1);if(nd0.val==nd1.val){flag=false;sum+=nd0.val;}else if(nd0.val<nd1.val){/// nd1if(nd1.unique==false) flag=false;sum+=nd1.val;}else if(nd0.val>nd1.val){/// nd0if(nd0.unique==false) flag=false;sum+=nd0.val;}}dp[now][0].val=sum;dp[now][0].unique=flag;vised[now][0]=true;return dp[now][0];}}int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);while(scanf("%d",&n)!=EOF&&n){init();cin>>name1;getID(name1);for(int i=0;i<n-1;i++){cin>>name1>>name2;int u=getID(name1);int v=getID(name2);vi[v].push_back(u);}bool flag=false;int ans=0;Node nd1=dfs(1,1);Node nd0=dfs(1,0);if(nd1.val==nd0.val){ans=nd1.val;flag=false;}else if(nd1.val<nd0.val){ans=nd0.val;flag=nd0.unique;}else if(nd1.val>nd0.val){ans=nd1.val;flag=nd1.unique;}printf("%d ",ans);if(flag) puts("Yes");else puts("No");}return 0;}
，累死累活不说，走马观花反而少了真实体验，
