补题：HNU暑期训练第三场-Charles in Charge（二分查找+dijstra（堆优化））

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题目链接：
Charles in Charge

题目大意：
给你n个点，m条边，找出满足点1到点n的路中不超过点1到点n最短路d（1+x%)经过的边的权值最大值最小，输出这个值

解题思路：
利用二分查找，二分这个最小值，然后利用dijstra算法去检验是否可以找到一条路满足经过的权值不大雨当前的限制，并且求出的最短路不超过d(1+x%),有的话，r=mid；无的话，l=mid+1

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>


using namespace std;

const long long MAXN = 1e5+7;
const long long MAXM = 1e5+7;
const long long INF = 0x7f7f7f7f3f3f3f3f;
#define ll long long

struct edge{
    int v;
    ll l;
    edge(){}
    edge(int _v, ll _l):v(_v),l(_l){} 
    bool operator <(const edge &e)const
    {
        return e.l < l;
    }
};

vector<edge> g[MAXN];

ll vis[MAXN],dist[MAXN];
ll n,m,x;

ll dijstra(ll limit){
    memset(vis,0,sizeof(vis));
    for(int i=1; i<=n; ++i) dist[i] = INF;

    priority_queue<edge> pq;
    dist[1] = 0;
    pq.push(edge(1, 0));
    while(!pq.empty()){
        edge tmp = pq.top();
        pq.pop();
        int u = tmp.v;
        if(vis[u]) continue;
        vis[u] = 1;
        for(int j=0; j<g[u].size(); ++j){
            int v = g[u][j].v;
            if(!vis[v] && g[u][j].l <= limit){
                if(dist[v]>dist[u] + g[u][j].l){
                    dist[v] = dist[u] + g[u][j].l;
                    pq.push(edge(v,dist[v]));
                }
            }
        }
    }
    return dist[n];
}


int main(){
    cin>>n>>m>>x;
    for(int i=1; i<=m; ++i){
        int u,v;
        ll t;
        cin>>u>>v>>t;
        g[u].push_back(edge(v,t));
        g[v].push_back(edge(u,t));
    }
    ll d = dijstra(INF);
    d = d + d * x / 100;

    ll l=0, r = d+233;
    while(l<r){
        ll mid = l+ (r-l) / 2;
        ll ans = dijstra(mid);
        if(ans <= d) r = mid;
        else l = mid+1;
    }

    cout<<l<<endl;
    return 0;
}