分层图最短路

Telephone Lines

https://vjudge.net/contest/350871#problem/L

题意：有n个点m条边，每条边为双向且有一个权值，最多可以让k条边的权值变为0，现要最小化从n号点到1号点经过的边权的最大值

思路1：将原图看作k+1层，每层有n个点，$dis[u][j]$ 表示到第j层的第u号点的边权最大值，即为当有j条边的权值置0时到u号点权值的最大值，可用$dis[u][j]$更新$dis[v][j+1]$（其中u,v之间有边相连,边权为w)，即将(u,v)这条边的权值置0，用max($dis[u][j]$,$w$)更新$dis[v][j]$，即(u,v)这条边不置0的情况，类似SPFA跑最短路即可（注：这题卡dijkstra不卡SPFA)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define MAXN 1005
#define MAXM 10005
#define INF 0x3f3f3f3f
using namespace std;
int head[MAXN],tot;
struct edge
{
	int v,w,nxt;
}edg[MAXM << 1];
inline void addedg(int u,int v,int w)
{
	edg[tot].v = v;
	edg[tot].w = w;
	edg[tot].nxt = head[u];
	head[u] = tot++;
}
int n,m,dis[MAXN][MAXN],k,cnt[MAXN][MAXN],vis[MAXN][MAXN];
struct node
{
	int u,c;
	node(){}
	node(int u,int c):u(u),c(c){}
}nod;
queue<node> qu;
inline bool SPFA(int s)
{
	memset(dis,0x3f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	memset(cnt,0,sizeof(cnt));
	while(!qu.empty())
		qu.pop();
	dis[s][0] = 0;
	vis[s][0] = 1;
	qu.push(node(s,0));
	int u,v,c;
	while(!qu.empty())
	{
		nod = qu.front();
		qu.pop();
		u = nod.u,c = nod.c;
		vis[u][c] = 0;
		for(int i = head[u];i != -1;i = edg[i].nxt)
		{
			v = edg[i].v;
			if(max(dis[u][c],edg[i].w) < dis[v][c])
			{
				dis[v][c] = max(dis[u][c],edg[i].w);
				if(!vis[v][c])
				{
					vis[v][c] = 1;
					qu.push(node(v,c));
					if(++cnt[v][c] > n*(k+1))
						return false;
				}
			}
			if(c < k && dis[u][c] < dis[v][c+1])
			{
				dis[v][c+1] = dis[u][c];
				if(!vis[v][c+1])
				{
					vis[v][c+1] = 1;
					qu.push(node(v,c+1));
					if(++cnt[v][c+1] > n*(k+1))
						return false;
				}
			}
		}
	}
	return true;
}
inline void init()
{
	memset(head,-1,sizeof(int)*(n+1));
	tot = 0;
}
int main()
{
	while(~scanf("%d%d%d",&n,&m,&k))
	{
		int u,v,w;
		init();
		while(m--)
		{
			scanf("%d%d%d",&u,&v,&w);
			addedg(u,v,w);
			addedg(v,u,w);
		}
		SPFA(n);
		int minn = INF;
		for(int i = 0;i <= k;++i)
			minn = min(minn,dis[1][i]);
		if(minn == INF)
			minn = -1; 
		printf("%d\n",minn);
	}
	return 0;
}

思路2：
https://blog.youkuaiyun.com/xing_mo/article/details/103926607