【CF】174 Div.1 B Cow Program

思路是树形DP+状态压缩。其实仅有2个状态，奇数次来到x或者偶数次来到x。（因为对x的更新不同）。
同时开辟visit数组，解决环。注意，一旦遇到环结果就是-1。DP数组存放第奇数/偶数次来到x时，对y的改变两。

 /* 283B */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")
 
 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1
 
 const int maxn = 2e5+;
 int a[maxn];
 __int64 dp[maxn][];
 bool visit[maxn][];
 int n;
 
 __int64 dfs(int x, int d) {
     if (x<= || x>n)
         return ;
 
     if (dp[x][d] != )
         return dp[x][d];
 
     if (visit[x][d]) {
         dp[x][d] = -;
         return -;
     }
 
     visit[x][d] = true;
     int x_ = d ? x + a[x] : x - a[x];
     int d_ = !d;
     __int64 ret = dfs(x_, d_);
 
     ret = ret <  ? - : ret + a[x];
     dp[x][d] = ret;
 
     return ret;
 }
 
 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif
 
     __int64 x, y, tmp;
 
     scanf("%d", &n);
     rep(i, , n+)
         scanf("%d", &a[i]);
 
     memset(visit, false, sizeof(visit));
     dp[][] = -;
     dp[][] = ;
 
     rep(i, , n) {
         y = i;
         x = i + ;
         tmp = dfs(x, );
         y = tmp< ? - : y+tmp;
         printf("%I64d\n", y);
     }
 
     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif
 
     return ;
 }