标签:sig img back char etc scanf hide lse with
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。
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; }
标签:sig img back char etc scanf hide lse with
原文地址:https://www.cnblogs.com/ckxkexing/p/10348316.html