题意:给你n个站点,m条无向边,求一条从1号站点到n号站点的路径,上面有k条路径免费的最大路径最小是多少?
题解:看到最大路径最小应该要想到二分的,但是因为看到了k条路径免费,又往分层图那边想了呜呜。二分答案,然后dijkstra判定,如果,那条边大于mid,那么,那条边赋值为1,否则赋值为0,如果最短路小于K,说明这个答案是ok的,二分找即可。
#include <cstdio>
#include <queue>
#include <cstring>
#define ll long long
#define pi pair<int,int>
#define mk make_pair
#define pb push_back
using namespace std;
const int maxn = 1e3+10;
vector<pi>G[maxn];
int n,m,K;
int vis[maxn],d[maxn];
bool ok(int ans)
{
priority_queue<pi>q;
memset(d,0x3f,sizeof(d));
memset(vis,0,sizeof(vis));
q.push(mk(0,1));
d[1] = 0;
while(!q.empty())
{
int u = q.top().second;
q.pop();
if(vis[u])continue;
vis[u] = 1;
for(int i=0;i<G[u].size();i++)
{
int v = G[u][i].first;
int w = G[u][i].second;
if(w <= ans)w = 0;
else w = 1;
if(d[v] > d[u] + w)
{
d[v] = d[u] + w;
q.push(mk(-d[v],v));
}
}
}
if(d[n] <= K)return true;
else return false;
}
int main()
{
int mx=0;
scanf("%d%d%d",&n,&m,&K);
for(int i=1;i<=m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
G[u].pb(mk(v,w));
G[v].pb(mk(u,w));
mx = max(mx,w);
}
int l = 0,r = mx,ans = -1;
while(l <= r)
{
int mid = (l+r)/2;
if(ok(mid))r = mid - 1,ans = mid;
else l = mid + 1;
}
printf("%d\n",ans);
return 0;
}