【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\n", y);
}

#ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif

return 0;
}