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…

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…

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…

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…

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…

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…

题目要求1到n点的最大容量的增广路. 听说是最短路求的,然后乱搞就A了.. 大概能从Bellman-Ford的思想,dk[u]表示从源点出发经过最多k条边到达u点的最短路,上理解正确性. #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; #define INF (1<<30) #define MAXN 1111 #…

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,…

题目链接:http://poj.org/problem?id=1797 题目就是求所有可达路径的其中的最小值边权的最大值 即对于每一条能够到达的路径,其必然有其最小的承载(其实也就是他们自身的最大的承载): 我们要求的就是所有路径的最小承载中的最大值 代码: #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> using namespace std; #define…

解题思路:典型的Kruskal,不能用floyed(会超时),上代码: #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define inf 0x3f3f3f3f ; int father[maxn]; struct node{ int x, y, w; }p[maxn*maxn]; //第一次开maxn,RE了一次 int cmp(node A, node B)…