题目链接:http://poj.org/problem?id=1797 Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 39999   Accepted: 10515 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand…

题目电波: POJ--1797 Heavy Transportation n点m条边, 求1到n最短边最大的路径的最短边长度 改进dijikstra,dist[i]数组保存源点到i点的最短边最大的路径的最短边长度 #include<iostream> #include<cstring> #include<algorithm> #include<stdio.h> using namespace std; #define maxn 100010 #define…

题目:click here 题意: 有n个城市,m条道路,在每条道路上有一个承载量,现在要求从1到n城市最大承载量,而最大承载量就是从城市1到城市n所有通路上的最大承载量.分析: 其实这个求最大边可以近似于求最短路,只要修改下找最短路更新的条件就可以了. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #i…

Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K          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 real…

Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 53170   Accepted: 13544 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man…

F - Heavy Transportation Time Limit:3000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1797 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand busines…

Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 22440   Accepted: 5950 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man…

Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 21037   Accepted: 5569 Description Background  Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man…

题目大意: 给你以T, 代表T组测试数据,一个n代表有n个点, 一个m代表有m条边, 每条边有三个参数,a,b,c表示从a到b的这条路上最大的承受重量是c, 让你找出一条线路,要求出在这条线路上的最小承重, 在所有其他线路最大. 题目分析: 这里只要将spfa进行一下变形就可以解决这问题了. 首先 我们的dist数组,起点位置要初始化为 INF, 其他位置初始化为 0 然后我们更新 dist 数组, 结果输出 dist[n]就行了 为什么这样写: 因为我们每次要找 所有路径中的最大边的最小一个,…

传送门 1.最大生成树 可以求出最大生成树,其中权值最小的边即为答案. 2.最短路 只需改变spfa里面的松弛操作就可以求出答案. ——代码 #include <queue> #include <cstdio> #include <cstring> using namespace std; ; int T, n, m, cnt; int head[MAXN], next[MAXN * MAXN], to[MAXN * MAXN], val[MAXN * MAXN], d…