POJ 3013 Big Christmas Tree(最短Dijkstra+优先级队列优化,SPFA)
POJ 3013 Big Christmas Tree(最短路Dijkstra+优先队列优化,SPFA)
ACM
题目地址:POJ 3013
题意:
圣诞树是由n个节点和e个边构成的,点编号1-n。树根为编号1,选择一些边。使得全部节点构成一棵树。选择边的代价是(子孙的点的重量)×(这条边的价值)。
求代价最小多少。
分析:
单看每一个点被计算过的代价,非常明显就是从根到节点的边的价值。所以这是个简单的单源最短路问题。
只是坑点还是非常多的。
点的数量高达5w个,用矩阵存不行。仅仅能用边存。
还有路径和结果会超int。所以要提高INF的上限。(1<<16)*50000
就可以。
能够用Dijkstra+优先队列做,也能够用SPFA做,貌似SPFA会更快。我这里用的是Dijkstra,要1s多...回头要用SPFA做一遍。
用SPFA做了一遍发现也是1s多,看了是STL用多了 = =。
嘛。留个模板。
代码:
(Dijkstra+priority_queue)
/* * Author: illuz <iilluzen[at]gmail.com> * File: 3013.cpp * Create Date: 2014-07-27 09:54:35 * Descripton: dijkstra */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #include <vector> #include <queue> #define repf(i,a,b) for(int i=(a);i<=(b);i++) const int N = 50100; const long long INF = (long long)(1<<16)*N; struct Edge { int from, to; int dist; }; struct HeapNode { int d; int u; bool operator < (const HeapNode rhs) const { return d > rhs.d; } }; struct Dijkstra { int n, m; // number of nodes and edges vector<Edge> edges; vector<int> G[N]; // graph bool vis[N]; // visited?long long d[N]; // dis int p[N]; // prevent edge void init(int _n) { n = _n; } void relief() { for (int i = 0; i < n; i++) { G[i].clear(); } edges.clear(); } void AddEdge(int from, int to, int dist) { // if non-directed, add twice edges.push_back((Edge){from, to, dist}); m = edges.size(); G[from].push_back(m - 1); } void dijkstra(int s) { priority_queue<HeapNode> Q; for (int i = 0; i < n; i++) { d[i] = INF; vis[i] = 0; } d[s] = 0; Q.push((HeapNode){0, s}); while (!Q.empty()) { HeapNode x = Q.top(); Q.pop(); int u = x.u; if (vis[u]) { continue; } vis[u] = true; for (int i = 0; i < G[u].size(); i++) { // update the u's linking nodes Edge& e = edges[G[u][i]]; //ref for convenient if (d[e.to] > d[u] + e.dist) { d[e.to] = d[u] + e.dist; p[e.to] = G[u][i]; Q.push((HeapNode){d[e.to], e.to}); } } } } }; int t; int e, v, x, y, d, w[N]; int main() { scanf("%d", &t); Dijkstra di; while (t--) { scanf("%d%d", &v, &e); di.init(v); repf (i, 0, v - 1) { scanf("%d" ,&w[i]); } repf (i, 0, e - 1) { scanf("%d%d%d", &x, &y, &d); di.AddEdge(x - 1, y - 1, d); di.AddEdge(y - 1, x - 1, d); } di.dijkstra(0); long long ans = 0; bool ring = false; repf (i, 0, v - 1) { if (di.d[i] == INF) { ring = true; } ans += w[i] * di.d[i]; } if (ring) { cout << "No Answer" << endl; } else { cout << ans << endl; } if (t) // if not the last case di.relief(); } return 0; }
(SPFA)
/* * Author: illuz <iilluzen[at]gmail.com> * File: 3013_spfa.cpp * Create Date: 2014-07-27 15:44:45 * Descripton: spfa */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #include <vector> #include <queue> #define repf(i,a,b) for(int i=(a);i<=(b);i++) const int N = 50100; const long long INF = (long long)(1<<16)*N; struct Edge { int from, to; int spst; }; struct SPFA { int n, m; vector<Edge> edges; vector<int> G[N]; // the edges which from i bool vis[N]; long long d[N]; // sps int p[N]; // prevent void init(int _n) { n = _n; } void relief() { for (int i = 0; i < n; i++) G[i].clear(); edges.clear(); } void AddEdge(int from, int to, int spst) { // if non-sprected, add twice edges.push_back((Edge){from, to, spst}); m = edges.size(); G[from].push_back(m - 1); } void spfa(int s) { queue<int> Q; while (!Q.empty()) Q.pop(); for (int i = 0; i < n; i++) { d[i] = INF; vis[i] = 0; } d[s] = 0; vis[s] = 1; Q.push(s); while (!Q.empty()) { int u = Q.front(); Q.pop(); vis[u] = 0; for (int i = 0; i < G[u].size(); i++) { Edge& e = edges[G[u][i]]; if (d[e.to] > d[u] + e.spst) { d[e.to] = d[u] + e.spst; p[e.to] = G[u][i]; if (!vis[e.to]) { vis[e.to] = 1; Q.push(e.to); } } } } } }; int t; int e, v, x, y, d, w[N]; int main() { scanf("%d", &t); SPFA sp; while (t--) { scanf("%d%d", &v, &e); sp.init(v); repf (i, 0, v - 1) { scanf("%d" ,&w[i]); } repf (i, 0, e - 1) { scanf("%d%d%d", &x, &y, &d); sp.AddEdge(x - 1, y - 1, d); sp.AddEdge(y - 1, x - 1, d); } sp.spfa(0); long long ans = 0; bool ring = false; repf (i, 0, v - 1) { if (sp.d[i] == INF) { ring = true; } ans += w[i] * sp.d[i]; } if (ring) { cout << "No Answer" << endl; } else { cout << ans << endl; } if (t) // if not the last case sp.relief(); } return 0; }
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