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
Hint
In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.
第一个DP状态压缩的题，看了好久。希望下次做的时候可以懂一点点吧。
// Asimple #include <iostream> #include <algorithm> #include <cstdio> #include <cstdlib> #include <queue> #include <vector> #include <string> #include <cstring> #include <stack> #define INF 0x3f3f3f3f #define mod 2016 using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = (1<<15)+10; int n, m, T, len, cnt, num, ans, Max; int dp[maxn], t[maxn], pre[maxn], d[20], f[20]; char str[20][105]; void output(int x) { if( !x ) return ; output(x-(1<<pre[x])); printf("%s ", str[pre[x]]); } void input() { scanf("%d", &T); while( T -- ) { scanf("%d", &n); for(int i=0; i<n; i++) scanf("%s%d%d",str[i], &d[i], &f[i]); int bit = 1 << n; for(int i=1; i<=bit; i++) { dp[i] = INF; for(int j=n-1; j>=0; j--) { int te = 1<<j; if( !(i&te) ) continue; int sc = t[i-te]+f[j]-d[j]; if( sc<0 ) sc = 0; if( dp[i]>dp[i-te]+sc) { dp[i] = dp[i-te]+sc; t[i] = t[i-te]+f[j]; pre[i] = j; } } } printf("%d ", dp[bit-1]); output(bit-1); } } int main() { input(); return 0; }
感谢大神http://blog.csdn.net/xingyeyongheng/article/details/21742341