HDU 1595 find the longest of the shortest 删掉任意一条边的最长最短路

题目来源：HDU 1595 find the longest of the shortest

题意：n个点m条边的无向图 有一条路是不能走的 求从1到n最短路的最大值

思路：首先求出最短路 如果坏掉的那条路不在最短路上 那么最短路是不会变大的 所以枚举最短路上的n-1条边 每次把这条路设为正无穷 在做最短路 取最大值

一次Dijkstra的最短路的复杂度是mlogn n次最短路的复杂度是nmlogn

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 1010;
struct edge
{
	int u, v, w;
}a[maxn*maxn];
int n, m, num;
struct HeapNode
{
	int u, d;
	bool operator < (const HeapNode& rhs) const
	{
		return d > rhs.d;
	}
};
vector <edge> edges;
vector <int> G[maxn];
int d[maxn];
int p[maxn];
bool vis[maxn];
void Dijkstra(int s)
{
	for(int i = 0; i <= n; i++)
		d[i] = 999999999;
	d[1] = 0;
	memset(vis, false, sizeof(vis));
	if(s)
		memset(p, -1, sizeof(p));
	priority_queue <HeapNode> Q;
	HeapNode x;
	x.u = 1;
	x.d = 0;
	Q.push(x);
	while(!Q.empty())
	{
		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++)
		{
			edge e = edges[G[u][i]];
			int v = e.v;
			if(d[v] > x.d + e.w)
			{
				if(s)
					p[v] = G[u][i];
				d[v] = x.d + e.w;
				HeapNode xx;
				xx.u = v;
				xx.d = d[v];
				Q.push(xx);
			}
		}
	}
}
void AddEdge(int u, int v, int w)
{
	edge x;
	x.u = u;
	x.v = v;
	x.w = w;
	edges.push_back(x);
	x.u = v;
	x.v = u;
	edges.push_back(x);
	num = edges.size();
	G[u].push_back(num-2);
	G[v].push_back(num-1);
}
int main()
{
	while(scanf("%d %d", &n, &m) != EOF)
	{
		for(int i = 0; i <= n; i++)
			G[i].clear();
		edges.clear();
		for(int i = 0; i < m; i++)
		{
			scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].w);
			AddEdge(a[i].u, a[i].v, a[i].w);
		}
		Dijkstra(1);
		int ans = -1;
		int pos = p[n];
		
		while(pos != -1)
		{
			//printf("%d %d\n", pos, pos^1);
			int w1 = edges[pos].w;
			int w2 = edges[pos^1].w;
			
			edges[pos].w = 999999999;
			edges[pos^1].w = 999999999;
			Dijkstra(0);
			edges[pos].w = w1;
			edges[pos^1].w = w2;
			
			pos = p[edges[pos].u];
			
			if(d[n] == 999999999)
				continue;
			if(ans == -1 || ans < d[n])
				ans = d[n];
			
		}
		printf("%d\n", ans);
	}
	return 0;
}