P2157 [SDOI2009]学校食堂 状压DP

题意：　　

　　排队买饭，时间为前一个人和后一个人的异或和，每个人允许其后面$B[i]$ 个人先买到饭，问最少的总用时。

思路：

　　用$dp[i][j][k]$表示$1～i-1$已经买好饭了，第i个人后面买饭情况为$j$，最后一个打饭的是$i+k$。

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC                      \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define FT(A, B, C) for (int A = B; A <= C; ++A) //用来压行
#define REP(i, j, k) for (int i = j; i < k; ++i)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1000000007;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
/*-----------------------showtime----------------------*/
const int maxn = 1009;

int dp[maxn][1 << 8][20];
int A[maxn], B[maxn];
int cal(int q, int h) {
    if (q == 0) return 0;
    return A[q] ^ A[h];
}

void solve() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d%d", &A[i], &B[i]);
    memset(dp, inf, sizeof(dp));
    dp[1][0][7] = 0;

    for (int i = 1; i <= n; i++)
        for (int j = 0; j < (1 << 8); j++) {
            for (int k = -8; k <= 7; k++)
                if (dp[i][j][k + 8] < inf) {
                    if (j & 1)
                        dp[i + 1][j >> 1][k + 7] = min(dp[i + 1][j >> 1][k + 7], dp[i][j][k + 8]);
                    else {
                        int mx = inf;
                        for (int h = 0; h <= 7; h++)
                            if (!(j & (1 << h))) {
                                if (i + h > mx) break;
                                mx = min(mx, i + h + B[i + h]);
                                int tmp = dp[i][j][k + 8] + cal(i + k, i + h);
                                dp[i][j | (1 << h)][h + 8] = min(dp[i][j | (1 << h)][h + 8], tmp);
                            }
                    }
                }
        }
    int ans = inf;
    for (int i = 0; i <= 8; i++) ans = min(ans, dp[n + 1][0][i]);
    cout << ans << endl;
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        solve();
    }
    return 0;
}