zzulioj 2634: E（spfa限制边权+二分）

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;
}