大家好,又见面了,我是全栈君,今天给大家准备了Idea注册码。
The more, The Better
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5414 Accepted Submission(s): 3217
b 代表第 i 个城堡的宝物数量, b >= 0。当N = 0, M = 0输入结束。
3 2 0 1 0 2 0 3 7 4 2 2 0 1 0 4 2 1 7 1 7 6 2 2 0 0
5 13dp[i][j]表示以i为根节点j个子节点的最大值。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> #include<vector> typedef long long LL; using namespace std; const int maxn=220; int v[maxn]; int n,m; int dp[maxn][maxn]; vector<int>s[maxn]; void tree_dp(int n,int f) { int len=s[n].size(); dp[n][1]=v[n]; for(int i=0;i<len;i++) { if(f>1) tree_dp(s[n][i],f-1); for(int j=f;j>=1;j--) { for(int k=1;k<=j;k++) dp[n][j+1]=max(dp[n][j+1],dp[n][j+1-k]+dp[s[n][i]][k]); } } } int main() { int f; while(~scanf("%d%d",&n,&m)&&(n+m)) { v[0]=0; memset(dp,0,sizeof(dp)); for(int i=0;i<=n;i++) s[i].clear(); for(int i=1;i<=n;i++) { scanf("%d%d",&f,&v[i]); s[f].push_back(i); } tree_dp(0,m+1); printf("%d\n",dp[0][m+1]); } return 0; }Anniversary party
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4329 Accepted: 2463 Description
There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. The University has a hierarchical structure of employees. It means that the supervisor relation forms a tree rooted at the rector V. E. Tretyakov. In order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. The personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. Your task is to make a list of guests with the maximal possible sum of guests' conviviality ratings.
Input
Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. Each of the subsequent N lines contains the conviviality rating of the corresponding employee. Conviviality rating is an integer number in a range from -128 to 127. After that go N – 1 lines that describe a supervisor relation tree. Each line of the tree specification has the form:
L K
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line
0 0
Output
Output should contain the maximal sum of guests' ratings.
Sample Input
7 1 1 1 1 1 1 1 1 3 2 3 6 4 7 4 4 5 3 5 0 0Sample Output
5#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> typedef long long LL; using namespace std; const int maxn=6005; int dp[maxn][2],pre[maxn]; int visit[maxn],n; void tree_dp(int x) { visit[x]=1; for(int i=1;i<=n;i++) { // cout<<"111 "<<i<<endl; if(!visit[i]&&pre[i]==x) { tree_dp(i); dp[x][1]+=dp[i][0]; dp[x][0]+=max(dp[i][1],dp[i][0]); } } } int main() { while(~scanf("%d",&n)) { memset(dp,0,sizeof(dp)); memset(visit,0,sizeof(visit)); memset(pre,0,sizeof(pre)); for(int i=1;i<=n;i++) scanf("%d",&dp[i][1]); int x,y,root; while(~scanf("%d%d",&x,&y)&&(x+y)) { pre[x]=y; root=y; } while(pre[root]) root=pre[root]; // cout<<"fuck "<<root<<endl; tree_dp(root); printf("%d\n",max(dp[root][0],dp[root][1])); } return 0; }
发布者:全栈程序员-用户IM,转载请注明出处:https://javaforall.cn/117734.html原文链接:https://javaforall.cn
【正版授权,激活自己账号】: Jetbrains全家桶Ide使用,1年售后保障,每天仅需1毛
【官方授权 正版激活】: 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...