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Dijkstra(变形) POJ 1797 Heavy Transportation

2015-11-27 17:06 351 查看
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
;
int d
;
bool vis
;
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
);
}

return 0;
}


  
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