hdu1520：Anniversary party

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1520

参考思路：我用的是动态规划，dp[i][0]表示; 以i为根节点的子树，不选择第i个节点，可以得到的最大值，dp[i][1]表示：与dp[i][0]相似，只是选择第i个节点。状态转移方程式为（j为i的孩子节点）：

源代码：

#include <iostream>#include <vector>using namespace std;const int inf = 0x7fffffff;inline int myMax(int a, int b) {return a > b ? a : b;}vector<int> vi[6001];int rate[6001];bool isRoot[6001];int dp[6001][2];int root;int n;int dfs(int cur, int flag) {if(dp[cur][flag] != inf) {return dp[cur][flag];}if(vi[cur].size() == 0) {if(flag == 0)return dp[cur][0] = 0;elsereturn dp[cur][1] = rate[cur];}if(flag == 1) {vector<int>::iterator it = vi[cur].begin();dp[cur][1] = rate[cur];while(it != vi[cur].end()) {dp[cur][1] += dfs(*it, 0);it++;}return dp[cur][1];} else {vector<int>::iterator it = vi[cur].begin();dp[cur][0] = 0;while(it != vi[cur].end()) {dp[cur][0] += myMax(dfs(*it, 0), dfs(*it, 1));it++;}return dp[cur][0];}}int main(){while(scanf("%d", &n) == 1) {int i, j, k;for(i=1; i<=n; i++) {scanf("%d", &rate[i]);}for(i=1; i<=n; i++) {vi[i].clear();isRoot[i] = true;}int u, v;while(scanf("%d%d", &u, &v)) {if(u==0 && v==0)break;vi[v].push_back(u);isRoot[u] = false;}for(i=1; i<=n; i++) {if(isRoot[i]) {root= i;break;}}for(i=1; i<=n; i++) {dp[i][0] = dp[i][1] = inf;}printf("%d\n", myMax(dfs(root, 0), dfs(root, 1)));}return 0;}