【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+5;
int a[maxn];
__int64 dp[maxn][2];
bool visit[maxn][2];
int n;

__int64 dfs(int x, int d) {
    if (x<=0 || x>n)
        return 0;
    
    if (dp[x][d] != 0)
        return dp[x][d];
    
    if (visit[x][d]) {
        dp[x][d] = -1;
        return -1;
    }
    
    visit[x][d] = true;
    int x_ = d ? x + a[x] : x - a[x];
    int d_ = !d;
    __int64 ret = dfs(x_, d_);
    
    ret = ret < 0 ? -1 : 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, 2, n+1)
        scanf("%d", &a[i]);
    
    memset(visit, false, sizeof(visit));
    dp[1][1] = -1;
    dp[1][0] = 0;
    
    rep(i, 1, n) {
        y = i;
        x = i + 1;
        tmp = dfs(x, 0);
        y = tmp<0 ? -1 : y+tmp;
        printf("%I64d
", y);
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}