[kuangbin带你飞]专题四最短路练习 C

最新推荐文章于 2021-01-18 16:55:04 发布

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本文介绍了一个寻找从起点到终点的最大承载重量路径的问题。利用优先队列实现了一种改进的Dijkstra算法，该算法能有效找到有向图中任意两点间路径上的最小权重的最大值。

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http://poj.org/problem?id=1797
Description
Background
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.

Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo’s place) to crossing n (the customer’s place). You may assume that there is at least one path. All streets can be travelled in both directions.
Input

题意：

给定一个双向可达的有向图，求出1->n之间的所有可达路径，每条路径中的最小边的最大值。

tip:

上一个题是最长边最小，这个是最短边最大。。
最大堆，贪心的策略是每次的dis为前面路径的中最小值，但因为每次取出的都是最大值，故最后得到的一定正确。。
为毛线都是dijstra。。这个输出啊 excitd
\n\

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <deque>
#include <utility>
#include <functional>
const int maxn = 100010;
const double eps = 1e-6;
const int INF = (1<<30);
using namespace std;
int n,m,tot;
int dis[maxn],first[maxn];
typedef pair<int,int>pii;
priority_queue <pii> pq;

struct node{
    int v,w,next;
}edges[10*maxn];

void add(int u,int v,int w){
    edges[tot].v = v;edges[tot].w = w;edges[tot].next = first[u];first[u] = tot++;
    edges[tot].v = u;edges[tot].w = w;edges[tot].next = first[v];first[v] = tot++;
}

void dij(){
    while(!pq.empty()){
        int no = pq.top().second;
        if(no == n) return;
        pq.pop();

        for(int i = first[no]; i != -1;i = edges[i].next){
            if(min(edges[i].w ,dis[no]) > dis[edges[i].v]){
                dis[edges[i].v] = min(edges[i].w ,dis[no]);
                pq.push(make_pair(dis[edges[i].v],edges[i].v));
            }
        }
    }
}

void init(){
    tot = 0;
    memset(dis,0,sizeof(dis));
    memset(first,-1,sizeof(first));
    scanf("%d%d",&n,&m);
    for(int i = 0 ; i < m ; i++){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w);
    }

    while(!pq.empty()){
        pq.pop();
    }
    dis[1] = INF;
    pq.push(make_pair(0,1));
}

int main(){
    int T;
    scanf("%d",&T);
    for(int ca = 1 ;ca <= T ;ca++){
        init();
        dij();
        printf("Scenario #%d:\n%d\n\n",ca,dis[n]);
    }
}