Mr. Panda and crystal HDU-6007 short circuit + full backpack

Question: Question Link

Idea: it is not difficult to see that the synthesis of each gem consumes a certain amount of magic, each gem has a certain benefit, so as long as we know the minimum cost of each gem synthesis, this problem can be converted into a full backpack with the initial magic value. The minimum cost of each gem can be calculated by running Dijkstra once and then the shortest path length is expressed by the synthetic cost.

AC code:

  1 #include <iostream>  2 #include <cstdio>  3 #include <algorithm>  4 #include <cstring>  5 #include <vector>  6 #include <queue>  7   8 using namespace std;  9  10 const int maxn = 10000 + 5; 11  12 int vol, n, m, INF; 13  14 struct node { 15     int c, w; 16     vector<int> v; 17     vector<vector<pair<int, int> > > vec; 18     friend bool operator < (node a, node b) { 19         return a.w > b.w; 20     } 21 }gem[maxn]; 22  23 priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>> > q; 24  25 bool vis[maxn]; 26  27 void init() { 28     for(int i = 1; i <= n; ++i) { 29         for(int j = 0; j < gem[i].vec.size(); ++j) 30             gem[i].vec[j].clear(); 31         gem[i].vec.clear(); 32         gem[i].v.clear(); 33     } 34     while(!q.empty())  35         q.pop(); 36     memset(vis, false, sizeof(vis)); 37 } 38  39 bool get_sum(int id) { 40     int sum, _min = INF; 41     for(int i = 0; i < gem[id].vec.size(); ++i) { 42         sum = 0; 43         for(int j = 0; j < gem[id].vec[i].size(); ++j) { 44             sum += gem[gem[id].vec[i][j].first].c * gem[id].vec[i][j].second; 45             if(sum > INF) 46                 sum = INF; 47         } 48         _min = min(_min, sum); 49     } 50     if(_min < gem[id].c) { 51         gem[id].c = _min;  52         return true; 53     } 54     return false; 55 } 56  57 void dijkstra() { 58     while(!q.empty()) { 59         pair<int, int> P = q.top(); 60         q.pop(); 61         if(vis[P.second]) 62             continue; 63         vis[P.second] = true; 64         for(int i = 0; i < gem[P.second].v.size(); ++i) { 65             if(!vis[gem[P.second].v[i]] && get_sum(gem[P.second].v[i])) { 66                 q.push(make_pair(gem[gem[P.second].v[i]].c, gem[P.second].v[i])); 67             } 68         } 69     } 70 } 71  72 int dp[maxn]; 73  74 int main() 75 { 76     ios::sync_with_stdio(0); 77     cin.tie(0); 78  79     int T, t = 0; 80     cin >> T; 81     while(T--) { 82         cin >> vol >> n >> m; 83         INF = vol + 1; 84         init(); 85         int flag; 86         for(int i = 1; i <= n; ++i) { 87             cin >> flag; 88             if(flag) 89                 cin >> gem[i].c >> gem[i].w; 90             else { 91                 cin >> gem[i].w; 92                 gem[i].c = INF; 93             } 94         } 95         int id, num, c, nu; 96         vector<pair<int, int> > ope; 97         for(int i = 0; i < m; ++i) { 98             cin >> id >> num; 99             ope.clear();100             for(int j = 0; j < num; ++j) {101                 cin >> c >> nu;102                 ope.push_back(make_pair(c, nu));103                 gem[c].v.push_back(id);104             }105             gem[id].vec.push_back(ope);106         }107 108         for(int i = 1; i <= n; ++i) 109             if(gem[i].c < INF)110                 q.push(make_pair(gem[i].c, i));111 112         dijkstra();113 114         memset(dp, 0, sizeof(dp));115         for(int i = 1; i <= n; ++i) 116             for(int v = 0; v <= vol; ++v)117                 if(v >= gem[i].c)118                     dp[v] = max(dp[v], dp[v - gem[i].c] + gem[i].w);119     120         cout << "Case #" << ++t << ": " << dp[vol] << endl;121     }122     return 0;123 }

 

Mr. Panda and crystal HDU-6007 short circuit + full backpack