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，使得所有节点构成一棵树，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： 
单看每个点被计算过的代价，很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。

不过坑点还是很多的。 
点的数量高达5w个，用矩阵存不行，只能用边存。 
还有路径和结果会超int，所以要提高INF的上限，(1<<16)*50000即可。

可以用Dijkstra+优先队列做，也可以用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 edgesvector<Edge> edges;vector<int> G[N];// graphbool vis[N];// visited?long long d[N];// disint p[N];// prevent edgevoid 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 twiceedges.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 nodesEdge& e = edges[G[u][i]];//ref for convenientif (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 casedi.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 ibool vis[N];long long d[N];// spsint p[N];// preventvoid 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 twiceedges.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 casesp.relief();}return 0;}