P2157 [SDOI2009]学校食堂 状压DP

Posted 2021-02-07 ckxkexing

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 << "
";
#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 ‘
‘

#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;
}

View Code