POJ 1463 strategy IC game minimum point coverage, tree-like DP

/* It should be a minimum point coverage problem. You can split the points and use a bipartite graph. Here we use the tree-like DP to solve the leaf node I. Obviously: DP [I] [0] = 0, DP [I] [1] = 1 otherwise, DP [I] [0] = sum (DP [J] [1], J is the child of I ), DP [I] [1] = sum (min (DP [J] [0], DP [J] [1]) */# include <vector> # include <map> # include <set> # include <deque> # include <stack> # include <bitset> # include <algorithm> # include <functional> # include <numeric> # include <utility> # include <sstream> # include <iostream> # include <iomanip> # include <cstdio> # include <queue> # InCl Ude <cmath> # include <cstdlib> # include <cstring> # include <ctime> # include <string> using namespace STD; const int n = 1600; int N; struct edge {int V, next;} edge [2 * n]; int first [N], E; int DP [N] [2]; void addedge (int u, int v) {edge [e]. V = V; edge [e]. next = first [u]; first [u] = e ++; edge [e]. V = u; edge [e]. next = first [v]; first [v] = e ++;} void Init () {int X, num, Y; E = 0; For (INT I = 0; I <n; I ++) {first [I] =-1; DP [I] [0] = 0; DP [I] [1] = 1 ;} for (INT I = 0; I <n; I ++) {scanf ("% d :( % d)", & X, & num); For (Int J = 0; j <num; j ++) {scanf ("% d", & Y); addedge (x, y) ;}} int DFS (int x, int FA) {int flag = 1; for (int K = first [X]; k! =-1; k = edge [K]. Next) {If (edge [K]. V! = FA) {DFS (edge [K]. v, x); DP [x] [1] + = min (DP [edge [K]. v] [0], DP [edge [K]. v] [1]); DP [x] [0] + = DP [edge [K]. v] [1]; flag = 0 ;}} void solve () {DFS (0,-1); cout <min (DP [0] [0], DP [0] [1]) <Endl;} int main () {While (scanf ("% d", & N )! = EOF) {Init (); solve ();}}