P1462 通往奥格瑞玛的道路

标签：struct   memset   string   while   main   [1]   lse   答案   fine   

二分答案，大于二分值的边直接不走，看最后是否存在一条路径即可

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
#define debug(x) cerr << #x << "=" << x << endl;
const int MAXN = 10000 + 10;
const int MAXM = 500000 + 10;
int n,m,last[MAXN],tot,dist[MAXN],s,vis[MAXN],g[MAXN],f[MAXN],b;
queue <int> q;
struct Edge{
    int u,v,w,to;
    Edge(){}
    Edge(int u, int v, int w, int to) : u(u), v(v), w(w), to(to) {}
}e[MAXM];
inline void add(int u, int v, int w) {
    e[++tot] = Edge(u,v,w,last[u]);
    last[u] = tot;
}
bool SPFA(int s) {
    memset(dist,0x3f,sizeof(dist));
    memset(vis,0,sizeof(vis));
    dist[1] = 0;
    q.push(1);
    vis[1] = 1;
    while(!q.empty()) {
        int x = q.front();
        q.pop();
        vis[x] = 0;
        for(int i=last[x]; i; i=e[i].to) {
            int v = e[i].v;
            int w = e[i].w;
            if(dist[v] > dist[x] + w) {
                if(f[v] <= s) {
                    dist[v] = dist[x] + w;
                    if(!vis[v]) q.push(v), vis[v] = 1;
                }
            }
        }
    }
    if(dist[n] >= b) return false;
    else return true;
}
int main() {
    cin >> n >> m >> b;
    for(int i=1; i<=n; i++) {
        cin >> f[i];
        g[i] = f[i];
    }
    for(int i=1; i<=m; i++) {
        int a,b,c;
        cin >> a >> b >> c;
        add(a,b,c);
        add(b,a,c);
    }
    int l=1, r = n;
    sort(g+1,g+n+1);
    int ans=0;
    int mid;
    if(!SPFA(1<<30)) {
        cout << "AFK" << endl;
        return 0;
    }
    while(l <= r) {
        mid = l+r>>1;
        if(SPFA(g[mid])) {
            ans = g[mid];
            r = mid-1;
        } else {
            l = mid+1;
        }
    }
    cout << ans;
    return 0;
}

P1462 通往奥格瑞玛的道路

标签：struct   memset   string   while   main   [1]   lse   答案   fine   

原文地址：https://www.cnblogs.com/Zolrk/p/9918259.html