HDU - 1074 Doing Homework 状压DP

HDU - 1074

题意：

　　有n份作业，每份作业都有截止日期和完成这份作业的时间。超过截止日期一天多扣一分。问如何安排作业可以使的扣的分数最少。

思路：

　　由于n比较小，所以可以用状态压缩dp，2的15次 = 32 768。如何利用状态压缩呢？是这样的，每个数在二进制中都有15位，如果某个位值为1，表示当前这个作业已经完成。为0，则表示这个作业还没有做。所以假设当前状态为j，那么，$dp[j]$可以由比j少一个1的状态i转移过来。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#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 = 1e9+7;
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----------------------*/

            struct node{
                char str[200];
                int d,c;
            }a[30];
            const int maxn = 1e6+9;
            struct ode{
                int time,pre,cost,fa;
            }dp[maxn];
int main(){
            int n,T;  scanf("%d", &T);
            while(T--){
                scanf("%d", &n);

                for(int i=1; i<=n; i++)
                    scanf("%s%d%d", a[i].str, &a[i].d, &a[i].c);
                memset(dp,0,sizeof(dp));

                for(int j=1; j <(1<<n); j++){
                    dp[j].cost = inf;
                    for(int i=n; i>=1; i--){
                        int tmp = (1 << (i-1));

                        if((tmp & j) > 0){
                            int pp = j - tmp;
                            int fy = dp[pp].time  + a[i].c - a[i].d;
                            if(fy < 0) fy = 0;
                            if(dp[j].cost > dp[pp].cost + fy){
                                dp[j].cost = dp[pp].cost + fy;
                                dp[j].pre = i;
                                dp[j].fa = pp;
                                dp[j].time = dp[pp].time + a[i].c;
                            }
                        }
                    }
                }
                int q = (1<<n) - 1;
                //debug(q);
                printf("%d\n", dp[q].cost);
                stack <int>tmp;
                while(q > 0){
                    tmp.push(dp[q].pre);
                    q = dp[q].fa;
                }
                while(!tmp.empty()){
                    printf("%s\n", a[tmp.top()].str);
                    tmp.pop();
                }
            }
            return 0;
}