Mr. Panda and Crystal HDU - 6007 最短路+完全背包

题目：题目链接

思路：不难看出，合成每个宝石需要消耗一定的魔力值，每个宝石有一定的收益，所以只要我们知道每个宝石合成的最小花费，该题就可以转化为一个背包容量为初始魔力值的完全背包问题，每个宝石的最小花费可以用dijkstra跑一遍最短路算出，路径长度用合成花费表示。

AC代码：

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <queue>

 using namespace std;

 const int maxn =  + ;

 int vol, n, m, INF;

 struct node {
     int c, w;
     vector<int> v;
     vector<vector<pair<int, int> > > vec;
     friend bool operator < (node a, node b) {
         return a.w > b.w;
     }
 }gem[maxn];

 priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>> > q;

 bool vis[maxn];

 void init() {
     for(int i = ; i <= n; ++i) {
         for(int j = ; j < gem[i].vec.size(); ++j)
             gem[i].vec[j].clear();
         gem[i].vec.clear();
         gem[i].v.clear();
     }
     while(!q.empty())
         q.pop();
     memset(vis, false, sizeof(vis));
 }

 bool get_sum(int id) {
     int sum, _min = INF;
     for(int i = ; i < gem[id].vec.size(); ++i) {
         sum = ;
         for(int j = ; j < gem[id].vec[i].size(); ++j) {
             sum += gem[gem[id].vec[i][j].first].c * gem[id].vec[i][j].second;
             if(sum > INF)
                 sum = INF;
         }
         _min = min(_min, sum);
     }
     if(_min < gem[id].c) {
         gem[id].c = _min;
         return true;
     }
     return false;
 }

 void dijkstra() {
     while(!q.empty()) {
         pair<int, int> P = q.top();
         q.pop();
         if(vis[P.second])
             continue;
         vis[P.second] = true;
         for(int i = ; i < gem[P.second].v.size(); ++i) {
             if(!vis[gem[P.second].v[i]] && get_sum(gem[P.second].v[i])) {
                 q.push(make_pair(gem[gem[P.second].v[i]].c, gem[P.second].v[i]));
             }
         }
     }
 }

 int dp[maxn];

 int main()
 {
     ios::sync_with_stdio();
     cin.tie();

     int T, t = ;
     cin >> T;
     while(T--) {
         cin >> vol >> n >> m;
         INF = vol + ;
         init();
         int flag;
         for(int i = ; i <= n; ++i) {
             cin >> flag;
             if(flag)
                 cin >> gem[i].c >> gem[i].w;
             else {
                 cin >> gem[i].w;
                 gem[i].c = INF;
             }
         }
         int id, num, c, nu;
         vector<pair<int, int> > ope;
         for(int i = ; i < m; ++i) {
             cin >> id >> num;
             ope.clear();
             for(int j = ; j < num; ++j) {
                 cin >> c >> nu;
                 ope.push_back(make_pair(c, nu));
                 gem[c].v.push_back(id);
             }
             gem[id].vec.push_back(ope);
         }

         for(int i = ; i <= n; ++i)
             if(gem[i].c < INF)
                 q.push(make_pair(gem[i].c, i));

         dijkstra();

         memset(dp, , sizeof(dp));
         for(int i = ; i <= n; ++i)
             for(int v = ; v <= vol; ++v)
                 if(v >= gem[i].c)
                     dp[v] = max(dp[v], dp[v - gem[i].c] + gem[i].w);

         cout << "Case #" << ++t << ": " << dp[vol] << endl;
     }
     return ;
 }