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
;#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;}
每个人都有自己鲜明的主张和个性，
