Dijkstra(变形) POJ 1797 Heavy Transportation

 

题目传送门

题意:求1到n的最大载重量

分析:那么就是最大路上的最小的边权值,改变优先规则.

 

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

typedef long long ll;
const int N = 1e3 + 10;
const int E = 1e5 + 10;
const int INF = 0x3f3f3f3f;
struct	Edge	{
	int v, w, nex;
	Edge() {}
	Edge(int v, int w, int nex) : v (v), w (w), nex (nex) {}
	bool operator < (const Edge &r)	const	{
		return w < r.w;
	}
}edge[E];
int head[N];
int d[N];
bool vis[N];
int n, m, e;

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

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

void Dijkstra(int s)	{
	memset (d, 0, sizeof (d));
	memset (vis, false, sizeof (vis));
	d[s] = INF;
	priority_queue<Edge> que;	que.push (Edge (s, 0, 0));
	while (!que.empty ())	{
		int u = que.top ().v;	que.pop ();
		if (vis[u])	continue;
		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] < min (d[u], w))	{
				d[v] = min (d[u], w);	que.push (Edge (v, d[v], 0));
			}
		}
	}
}

int main(void)	{
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		init ();
		scanf ("%d%d", &n, &m);
		int mx = 0;
		for (int u, v, w, i=1; i<=m; ++i)	{
			scanf ("%d%d%d", &u, &v, &w);
			if (n == 1 && u == 1 && v == 1)	mx = max (mx, w);
			add_edge (u, v, w);
			add_edge (v, u, w);
		}
		printf ("Scenario #%d:\n", ++cas);
		if (n == 1)	{
			printf ("%d\n", mx);	continue;
		}
		Dijkstra (1);
		printf ("%d\n\n", d[n]);
	}

	return 0;
}