SPFA/Dijkstra POJ 3013 Big Christmas Tree

题目传送门

题意：找一棵树使得造价最少，造价为每个点的子节点造价和*边的造价和

分析：最短路跑出1根节点到每个点的最短边权值，然后每个点的权值*最短边距和就是答案，注意INF开足够大，n<=1特判。Dijkstra 和 SPFA都行

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;

typedef long long ll;
const int N = 5e4 + 10;
const ll INF = (1ll << 16) * 50000;
struct Edge	{
	int v, w, nex;
	Edge (int _v = 0, int _w = 0) : v (_v), w (_w) {}
	bool operator < (const Edge &r) const {
		return w > r.w;
	}
}edge[N*2];
ll d[N];
bool vis[N];
int head[N];
int c[N];
int cnt[N];
int n, m, e;

void init(void)	{
	memset (head, -1, sizeof (head));
	e = 0;
}

bool SPFA(int s)	{
	queue<int> Q;
	memset (vis, false, sizeof (vis));
	memset (cnt, 0, sizeof (cnt));
	d[s] = 0;	cnt[s] = 1;	vis[s] = true;
	Q.push (s);
	while (!Q.empty ())	{
		int u = Q.front ();	Q.pop ();
		vis[u] = false;
		for (int i=head[u]; ~i; i=edge[i].nex)	{
			int v = edge[i].v, w = edge[i].w;
			if (d[v] > d[u] + w)	{
				d[v] = d[u] + w;
				if (!vis[v])	{
					vis[v] = true;	Q.push (v);
					if (++cnt[v] > n)	return true;
				}
			}
		}
	}
	return false;
}

void Dijkstra(int s)	{
	priority_queue<Edge> Q;
	memset (vis, false, sizeof (vis));
	d[s] = 0;	Q.push (Edge (s, 0));
	while (!Q.empty ())	{
		int u = Q.top ().v;	Q.pop ();
		vis[u] = true;
		for (int i=head[u]; ~i; i=edge[i].nex)	{
			int v = edge[i].v, w = edge[i].w;
			if (!vis[v] && d[v] > d[u] + w)	{
				d[v] = d[u] + w;	Q.push (Edge (v, d[v]));
			}
		}
	}
}

void add_edge(int u, int v, int w)	{
	edge[e].v = v, edge[e].w = w;
	edge[e].nex = head[u];	head[u] = e++;
}

int main(void)	{
	int T;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d%d", &n, &m);
		for (int i=1; i<=n; ++i)	{
			scanf ("%d", &c[i]);	d[i] = INF;
		}
		init ();
		for (int u, v, w, i=1; i<=m; ++i)	{
			scanf ("%d%d%d", &u, &v, &w);
			add_edge (u, v, w);	add_edge (v, u, w);
		}
		if (n <= 1)	{
			puts ("0");	continue;
		}
		Dijkstra (1);
		// bool flag = SPFA (1);
		// if (flag)	{
		// 	puts ("No Answer");	continue;
		// }
		ll ans = 0;
		for (int i=1; i<=n; ++i)	{
			if (d[i] == INF)	{
				ans = -1;	break;
			}
			ans += d[i] * c[i];
		}
		if (ans == -1)	puts ("No Answer");
		else	printf ("%I64d\n", ans);
	}

	return 0;
}