算法问题实战策略 NTHLON

地址 https://algospot.com/judge/problem/read/NTHLON

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

 using namespace std;

 int V;
 //图的邻接表表示法 保存成对(连接的顶点序号 边线加权值)
 vector<pair<int, int>> adj[];
 const int START = ;
 const int INF = 0x3f3f3f3f;

 vector<int> dijkstra(int src)
 {
     vector<int > dist(V, INF);
     dist[src] = ;
     priority_queue<pair<int, int>> pq;
     pq.push(make_pair(, src));

     while (!pq.empty()) {
         int cost = -pq.top().first;
         int here = pq.top().second;
         pq.pop();

         if (dist[here] < cost) continue;

         for (int i = ; i < adj[here].size(); ++i) {
             int there = adj[here][i].first;
             int nextDist = cost + adj[here][i].second;

             if (dist[there] > nextDist) {
                 dist[there] = nextDist;
                 pq.push(make_pair(-nextDist, there));
             }
         }
     }

     return dist;
 }

 //==========================================================

 //返回(A国选手的预想耗时-B国家选手的预想耗时)的顶点序号
 int vertex(int delta) {
     return delta + ;
 }

 //a[]=A国选手的各项预想耗时
 //b[]=B国选手的各项预想耗时
 int solve(const vector<int>& a, const vector<int>& b) {
     //生成图结构
     V = ;
     for (int i = ; i < V; i++) adj[i].clear();
     for (int i = ; i < a.size(); i++) {
         int delta = a[i] - b[i];
         adj[START].push_back(make_pair(vertex(delta), a[i]));
     }

     //当前耗时之差
     for (int delta = -; delta <= ; ++delta) {
         //若添加第i个比赛项目
         for (int i = ; i < a.size(); ++i) {
             int next = delta + a[i] - b[i];
             //没有必要生成耗时之差超过200的顶点
             if (abs(next) > ) continue;
             adj[vertex(delta)].push_back(make_pair(vertex(next), a[i]));
         }
     }

     vector<int> shortest = dijkstra(START);
     int ret = shortest[vertex()];
     if (ret == INF) return -;
     return ret;
 }

 int main()
 {
     int n,m;
     cin >> n;
     while (n--) {
         cin >> m;
         vector<int> va;
         vector<int> vb;
         for (int i = ; i < m; i++) {
             int a, b;
             cin >> a >> b;
             va.push_back(a);
             vb.push_back(b);
         }
         int ret = solve(va, vb);
         if (ret != -) {
             cout << ret << endl;
         }
         else {
             cout << "IMPOSSIBLE" << endl;
         }
     }

     return ;
 }