9899828 2013-12-27 16:42:37 Accepted 1596 3312MS 6668K 711 B C++ 泽泽 floyed暴力 #include<cstdio> #include<cstring> ][]; void floyed(int n) { int k,j,i; ;k<=n;k++) { ;i<=n;i++) { ) { ;j<=n;j++) { if(g[i][k]*g[k][j]>g[i][j]) g[i][j]=g[j…

题目传送:http://acm.hdu.edu.cn/showproblem.php?pid=1596 分析:Dijkstra变体,最短路径判断计算方式:Safe(P) = s(e1)*s(e2)…*s(ek) e1,e2,ek是P 上的边. 判断方式:lowcost[pre]*cost[pre][j]>lowcost[j] 代码: //2013-10-30 10:54:06 Accepted 1596 1484MS 9788K 1836 B C++ 空信高手 #include <iostre…

Problem Description XX星球有很多城市,每个城市之间有一条或多条飞行通道,但是并不是所有的路都是很安全的,每一条路有一个安全系数s,s是在 0 和 1 间的实数(包括0,1),一条从u 到 v 的通道P 的安全度为Safe(P) = s(e1)*s(e2)…*s(ek) e1,e2,ek是P 上的边 ,现在8600 想出去旅游,面对这这么多的路,他想找一条最安全的路.但是8600 的数学不好,想请你帮忙 ^_^   Input 输入包括多个测试实例,每个实例包括:第一行:n.…

find the safest road Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9122    Accepted Submission(s): 3213 Problem Description XX星球有很多城市,每个城市之间有一条或多条飞行通道,但是并不是所有的路都是很安全的,每一条路有一个安全系数s,s是在 0 和 1 间…

find the safest road Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9033    Accepted Submission(s): 3173 Problem Description XX星球有非常多城市,每一个城市之间有一条或多条飞行通道,可是并非全部的路都是非常安全的,每一条路有一个安全系数s,s是在 0 和…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…