HDU 1074-doing Homework (pressure DP)

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Did a long time. Start with a search, write, write, think 15! How to prune the factorial again seems to be too much. In particular, when the dictionary sequence is not good processing AH later asked the Flying God is like pressure dp. Sad didn't know what it was like to be a pressure.

Test instructions: There are n homework to give each of the time and the completion of the work required to complete the operation of the optimal order, so that the minimum deduction (over the period to deduct points)

Idea: The completion of each job to see the status of 2, binary from right to left once the corresponding Job 1 2 ... n, the corresponding bit is 1 to indicate that the corresponding job is completed, and vice versa. So the general state has 2^n-1 species,

Dp[2^n-1] On behalf of all the work completed, Dp[0] on behalf of the work has not been completed. The DP array is open to the struct type, Dp[i] saves the current subtraction, the prefix, and the current time. Recursion from 0. Recursive printing. Most of these operations involve in-place operations, at first glance is very mindless, look at the definition of bit operations just fine

#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include < string> #include <cctype> #include <vector> #include <cstdio> #include <cmath> #include < queue> #include <stack> #include <map> #include <set> #define MAXN 1<<16#define _ll __int64# Define ll long long#define INF 0x3f3f3f3f#define Mod 100000000#define pp pair<int,int> #define ull unsigned long long Using namespace Std;struct node{int cost,re,pre;} Dp[maxn];char s[16][106];int n,c[16],d[16];bool vis[maxn];void init () {Dp[0].cost=0;dp[0].pre=-1;dp[0].re=0;memset ( Vis,0,sizeof (Vis));} void output (int sb) {int cur=dp[sb].pre^sb,cnt=0;while (cur) {cnt++;cur>>=1;} if (dp[sb].pre!=0) output (dp[sb].pre);p rintf ("%s\n", s[cnt]);} void Solve () {int tot= (1<<n) -1;for (int i=0;i<tot;i++) {for (int j=1;j<=n;j++) {int cur=1<< (j-1); Cur&i)) {int Tem=cur|i;dp[tem].cost=dp[i].cost+c[j];int Reduce=dp[tem].cost<d[j]?0:dp[tem].cosT-d[j];reduce+=dp[i].re;if (vis[tem]&&reduce<dp[tem].re) {dp[tem].re=reduce;dp[tem].pre=i;} if (!vis[tem]) {vis[tem]=1;dp[tem].re=reduce;dp[tem].pre=i;}}}} printf ("%d\n", dp[tot].re); output (tot);} int main () {int t;scanf ("%d", &t), while (t--) {scanf ("%d", &n), for (int i=1;i<=n;i++) scanf ("%s%d%d", s[i], &d[i],&c[i]); init (); Solve ();} return 0;}

HDU 1074-doing Homework (pressure DP)