POJ - 2342 Anniversary party 树形DP

来源：互联网发布：网络平台贷款逾期编辑：程序博客网时间：2024/05/18 01:59

题目大意：公司要开年会，要邀请员工，每个员工都有其对应的欢乐值。现要求在员工何其直属上司不能同时邀请的情况下，使得欢乐值最大

解题思路：设dp[i][1]表示邀请第i个人的情况，dp[i][0]表示没有邀请第i个人
那么dp[i][j] += sum(dp[j][0])
dp[i][0] = sum( max(dp[j][0],dp[j][1]))
dp[i][1]初始化为happy[i]，dp[i][0]初始化为0，这样就可以做了

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;#define maxn 6010vector<int> tree[maxn];int N, happy[maxn], dp[maxn][2], vis[maxn];void init() {    for(int i = 1; i <= N; i++) {        scanf("%d", &happy[i]);        vis[i] = 0;        tree[i].clear();    }    int L, K;    while(scanf("%d%d", &L, &K) != EOF && L + K) {        tree[K].push_back(L);        vis[L] = 1;    }    tree[0].clear();    for(int i = 1; i <= N; i++)        if(!vis[i])            tree[0].push_back(i);}void dfs(int cur) {    dp[cur][1] = happy[cur];    dp[cur][0] = 0;    for(int i = 0; i < tree[cur].size(); i++) {        dfs(tree[cur][i]);        dp[cur][1] += dp[tree[cur][i]][0];        dp[cur][0] += max(dp[tree[cur][i]][0], dp[tree[cur][0]][1]);    }}void solve() {    int ans = 0;    for(int i = 0; i < tree[0].size(); i++) {        dfs(tree[0][i]);        ans += max(dp[tree[0][i]][0], dp[tree[0][i]][1]);    }    printf("%d\n", ans);}int main() {    while(scanf("%d", &N) != EOF) {        init();        solve();    }    return 0;}

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