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;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = ;
const double esp = 1e-;
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=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
/*-----------------------showtime----------------------*/
            const int maxn = ;

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

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

                for(int i=; i<=n; i++)  for(int j=; j<(<<); j++){
                    for(int k=-; k<=; k++) if(dp[i][j][k+] < inf){
                        if(j & ) dp[i+][j>>][k+] = min(dp[i+][j>>][k+], dp[i][j][k+]);
                        else {
                            int mx = inf;
                            for(int h = ; h<=; h++) if(! (j & (<<h))){
                                if(i + h > mx) break;
                                mx = min(mx, i + h + B[i+h]);
                                int tmp = dp[i][j][k + ] + cal(i + k, i + h);
                                dp[i][j | ( << h)][h + ] = min(dp[i][j | ( << h)][h + ], tmp);
                            }
                        }
                    }
                }
                int ans = inf;
                for(int i=; i<=; i++) ans = min(ans, dp[n+][][i]);
                cout<<ans<<endl;
            }
int main(){
            int T;  scanf("%d", &T);
            while(T--){
                solve();
            }
            return ;
}