G. Training Camp
64-bit integer IO format: %I64d Java class name: (Any)
Montaser is planning to train very hard for ACM JCPC 2015; he has prepared a list with n topics
to study in the next n days, one topic every day. Montaser knows that some topics depend on other topics, so he asked coach Fegla and got a list
of m constraints on the order in which he should study these topics. Also, coach Fegla told him that when he studies topic x on the kth day (1 ≤ k ≤ n), his level will
increase by k*Wx, where Wx is a weight for topic x, representing how hard it is. Given the list of topics, the weight of each topic, and the list of constrains, can you tell Montaser
what is the maximum level he can reach in these n days? He is currently at level 0.
Input
The first line of input contains one integer T representing the number of test cases (1 ≤ T ≤ 128).
The first line of each test case contains two integers: n and m (1 ≤ n ≤ 18).
The next n lines, each contains the title of one of the topics followed by a space, then an integer
W that represents the weight of this topic (1 ≤ W ≤ 100).
The next m lines are of the form: Topic 1 --> Topic 2, which means that Topic 1 must be studied
before Topic 2.
Titles contain only English letters and spaces (no more than 40 characters).
Test cases are separated by a blank line.
Output
For each test case, print the maximum level that Montaser can reach.
Sample Input
1
3 2
Implementation 3
Dynamic Programming 10
Greedy 7
Greedy --> Dynamic Programming
Implementation --> Dynamic Programming
Sample Output
47
解题：状压动规
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 19; 4 int n,m,w[maxn],son[maxn],fa[maxn]; 5 vector<int>g[maxn]; 6 vector<int>o[maxn]; 7 unordered_map<string,int>ump; 8 void read(char *str,int pos){ 9 int len = strlen(str)-1,ret = 0; 10 while(len >= 0 && isdigit(str[len])) --len; 11 char topic[200]; 12 strncpy(topic,str,len); 13 topic[len] = '\0'; 14 for(int i = len + 1; str[i]; ++i) 15 ret = ret*10 + str[i] - '0'; 16 ump[topic] = pos; 17 w[pos] = ret; 18 } 19 void read2(char *str,int &a,int &b){ 20 int len = strlen(str) - 1; 21 while(len >= 0 && str[len] != '>') --len; 22 char topic[200]; 23 strncpy(topic,str,len-3); 24 topic[len-3] = '\0'; 25 a = ump[topic]; 26 strcpy(topic,str + len + 2); 27 b = ump[topic]; 28 } 29 int dfs(int u,vector<int>(&h)[maxn]){ 30 int ret = 0; 31 for(int i = h[u].size()-1; i >= 0; --i){ 32 ret |= (1<<h[u][i]); 33 ret |= dfs(h[u][i],h); 34 } 35 return ret; 36 } 37 int dp[1<<maxn],day[1<<maxn]; 38 int calc(int st){ 39 if(day[st] != -1) return day[st]; 40 int x = st; 41 st = (st&0x55555555) + ((st>>1)&0x55555555); 42 st = (st&0x33333333) + ((st>>2)&0x33333333); 43 st = (st&0x0F0F0F0F) + ((st>>4)&0x0F0F0F0F); 44 st = (st&0x00FF00FF) + ((st>>8)&0x00FF00FF); 45 st = (st&0x0000FFFF) + ((st>>16)&0x0000FFFF); 46 return day[x] = st; 47 } 48 int main(){ 49 int kase,a,b; 50 char str[200]; 51 scanf("%d",&kase); 52 while(kase--){ 53 scanf("%d%d",&n,&m); 54 ump.clear(); 55 getchar(); 56 for(int i = 0; i < n; ++i){ 57 g[i].clear(); 58 o[i].clear(); 59 gets(str); 60 read(str,i); 61 fa[i] = son[i] = 0; 62 } 63 for(int i = 0; i < m; ++i){ 64 gets(str); 65 read2(str,a,b); 66 g[a].push_back(b); 67 o[b].push_back(a); 68 } 69 memset(dp,0,sizeof dp); 70 memset(day,-1,sizeof day); 71 for(int i = 0; i < n; ++i){ 72 son[i] = dfs(i,g); 73 fa[i] = dfs(i,o); 74 } 75 m = (1<<n); 76 for(int i = 0; i < m; ++i){ 77 for(int j = 0; j < n; ++j){ 78 if(((i>>j)&1)) continue; 79 if(!(son[j]&i) && (fa[j]&i) == fa[j]) 80 dp[i|(1<<j)] = max(dp[i|(1<<j)],dp[i] + (calc(i) + 1)*w[j]); 81 } 82 } 83 printf("%d\n",dp[m-1]); 84 } 85 return 0; 86 }
