HDOJ 1074 Doing Homework

题意：有n门课的作业，每门课的作业都有必须花费时间以及截止期限，到截止期限还没做完的作业每天要扣一分，判断怎么安排作业顺序使得扣分最少

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1074

思路：n的范围很小(n<15)，容易想到是状压dp，在状态转移的过程中记录前一个状态的位置，最后利用stack还原顺序即可

注意点：无

以下为AC代码：

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor136697712015-05-11 10:34:04Accepted107415MS2648K3720 BG++luminous11

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <cstring>#include <cstdio>#include <string>#include <vector>#include <deque>#include <list>#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <cctype>#include <climits>#include <iomanip>#include <cstdlib>#include <algorithm>//#include <unordered_map>//#include <unordered_set>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define RDI(a) scanf ( "%d", &a )#define RDII(a, b) scanf ( "%d%d", &a, &b )#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )#define RS(s) scanf ( "%s", s )#define PI(a) printf ( "%d", a )#define PIL(a) printf ( "%d\n", a )#define PII(a,b) printf ( "%d %d", a, b )#define PIIL(a,b) printf ( "%d %d\n", a, b )#define PIII(a,b,c) printf ( "%d %d %d", a, b, c )#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )#define PL() printf ( "\n" )#define PSL(s) printf ( "%s\n", s )#define rep(i,m,n) for ( int i = m; i <  n; i ++ )#define REP(i,m,n) for ( int i = m; i <= n; i ++ )#define dep(i,m,n) for ( int i = m; i >  n; i -- )#define DEP(i,m,n) for ( int i = m; i >= n; i -- )#define repi(i,m,n,k) for ( int i = m; i <  n; i += k )#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )#define depi(i,m,n,k) for ( int i = m; i >  n; i += k )#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )#define READ(f) freopen(f, "r", stdin)#define WRITE(f) freopen(f, "w", stdout)using namespace std;const double pi = acos(-1);template <class T>inline bool RD ( T &ret ){    char c;    int sgn;    if ( c = getchar(), c ==EOF )return 0; //EOF    while ( c != '-' && ( c < '0' || c > '9' ) ) c = getchar();    sgn = ( c == '-' ) ? -1 : 1;    ret = ( c == '-' ) ? 0 : ( c - '0' );    while ( c = getchar() , c >= '0' && c <= '9' ) ret = ret * 10 + ( c - '0' );    ret *= sgn;    return 1;}inline void PD ( int x ){    if ( x > 9 ) PD ( x / 10 );    putchar ( x % 10 + '0' );}const double eps = 1e-10;const int inf = 1<<30;const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };struct node{    string str;    int dl;    int u;    void input ()    {        cin >> str >> dl >> u;    }}data[50];struct temp{    int now, pre;    int val, date;}dp[1<<16];int tmp;int n;int main(){    //ios::sync_with_stdio( false );    int k;    RDI ( k );    while ( k -- ){        clr ( dp, 0 );        RDI ( n );        rep ( i, 0, n ){            data[i].input();        }        int tmp = 1 << n;        rep ( s, 1, tmp ){            dp[s].val = inf;            DEP ( i, n - 1, 0 ){                int tem = 1 << i;                if ( s & tem ){                    int past = s - tem;                    int st = dp[past].date + data[i].u - data[i].dl;                    if ( st < 0 ) st = 0;                    if ( st + dp[past].val < dp[s].val ){                        dp[s].val = st + dp[past].val;                        dp[s].now = i;                        dp[s].pre = past;                        dp[s].date = dp[past].date + data[i].u;                    }                }            }        }        stack<int> S;        int tem = tmp - 1;        cout << dp[tem].val << endl;        while ( tem ){            S.push( dp[tem].now );            tem = dp[tem].pre;        }        while ( ! S.empty() ){            cout << data[S.top()].str << endl;            S.pop();        }    }    return 0;}