InputThe input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject's name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject's homework).
Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.
OutputFor each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.
Sample Input
2 3 Computer 3 3 English 20 1 Math 3 2 3 Computer 3 3 English 6 3 Math 6 3
Sample Output
2 Computer Math English 3 Computer English Math
DP姿势水平还是有待提高啊...
对于dp[x]，x的第i位为1则表示第i个作业已经完成，对于每个x枚举非零位j，如果dp[x] + (int)max(1ll*0,sumc - d[j])<dp[(1<<j)|x],说明从状态x完成j后到达
(1<<j)|x更优，则更新。维护del数组记录每个状态最近完成的作业是哪一个，然后通过dfs输出最优状态的完成顺序。
1 #include <cstdio> 2 #include <iostream> 3 #include <cmath> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <cstring> 7 #define N 20 8 #define pow2(_) (1<<_) 9 #define inf 0x3f3f3f3f 10 #define D(_) cout << #_ << " " << _ << endl; 11 using namespace std; 12 int T,n,d[N],c[N]; 13 char s[N][105]; 14 int dp[pow2(N)],del[pow2(N)]; 15 typedef long long ll; 16 17 void output(int n){ 18 if (n == pow2(del[n])) { 19 printf("%s ",s[del[n]]); 20 return; 21 } 22 output(n^pow2(del[n])); 23 printf("%s ",s[del[n]]); 24 } 25 26 int main(){ 27 scanf("%d",&T); 28 while(T--){ 29 scanf("%d",&n); 30 for (int i = 0;i < n;++i){ 31 scanf("%s %d %d",&s[i],&d[i],&c[i]); 32 } 33 memset(dp,0x3f3f3f3f,sizeof(dp)); 34 dp[0] = 0; 35 int maxn = (1<<n)-1; 36 for (int i = 0;i < maxn;++i){ 37 for (int j = 0;j < n;++j){ 38 if (pow2(j)&i) continue; 39 ll sumc = c[j]; 40 for (int k = 0;k < n;++k){ 41 if (pow2(k)&i) sumc += c[k]; 42 } 43 int tmp = dp[pow2(j)|i]; 44 dp[pow2(j)|i] = min(dp[pow2(j)|i],dp[i] + (int)max(1ll*0,sumc - d[j])); 45 if (tmp != dp[pow2(j)|i]) del[pow2(j)|i] = j; 46 } 47 } 48 49 printf("%d ",dp[pow2(n)-1]); 50 output(pow2(n)-1); 51 } 52 }