传送门
这道题是对这个全局最优的边权进行二分,我们知道这条路上的其他边的容量肯定都大于等于这个值。所以,就可以根据这个来限制当前的图,然后再判断当前图的可行性进行二分。
#include<cstdio>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<math.h>
#include<climits>
#include<set>
#include<sstream>
#include<time.h>
#include<iomanip>
#define debug(x) cout <<#x<<" = "<<x<<endl
#define debug2(x, y) cout<<#x<<" = "<<x<<", "<<#y<<" = "<<y<<endl
#define gg cout <<"---------------QAQ---------------"<<endl
#define fi first
#define SZ(x) (int)x.size()
#define se second
#define pb push_back
#define MEM(a) memset(a, 0, sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define PI 3.14
#define endl "\n"
#define eps 1e-8
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<pii, ll> Pii;
template<class T> inline void read(T &x){
x=0; char c=getchar(); int f=1;
while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
const int N = 2e5+10, maxn = 1e6+10;
void FAST(){ios::sync_with_stdio(false);cin.tie(nullptr); cout.tie(nullptr);}
const ll mod = 1e9+7;
int n, m, k;
struct edge
{
int u, v, w;
};
vector<edge> edges;
vector<int> e[N];
bool vis[N];
int d[N];
void add(int u, int v, int w)
{
edges.pb(edge{u, v, w});
int m = SZ(edges);
e[u].pb(m-1);
}
bool spfa(int limit)
{
queue<int> q;
MEM(vis);
memset(d, 0x3f, sizeof(d));
d[1] = 0;
q.push(1);
vis[1] = 1;
while(!q.empty())
{
int u = q.front(); q.pop();
vis[u] = 0;
for(int i = 0; i < SZ(e[u]); ++i) {
edge x = edges[e[u][i]];
int v = x.v, w = x.w;
int f = (w>limit);
if(d[v] > d[u]+f)
{
d[v] = d[u]+f;
if(!vis[v]) {
q.push(v);
vis[v] = 1;
}
}
}
}
return d[n]<=k;
}
void solve()
{
scanf("%d%d%d", &n, &m, &k);
while(m--)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
int l = 0, r = inf, ans = inf;
while(l <= r)
{
int m = (l+r)>>1;
if(spfa(m))
{
ans = min(ans, m);
r = m-1;
}
else l = m+1;
}
if(ans == inf) ans = -1;
printf("%d\n", ans);
}
int main()
{
// FAST();
// init();
// int _;scanf("%d", &_); while(_--)
// while(scanf("%d", &n)&&n)
// while(scanf("%d%d%d", &n, &m, &k)&&(n+m+k))
// for(ll i = 1; i <= _; ++i)
solve();
return 0;
}