#include<cstdio> #include<cstring> #define inf 0xffffff ][]; int ans; void prim(int n) { ],used[],i,j,k,min,closet[]; memset(used,,sizeof(used)); ;i<=n;i++) lowcost[i]=g[i][],closet[i]=; used[]=; ;i<n;i++) { j=; min=inf; ;k<=n;k++) {…

继续畅通工程 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10967    Accepted Submission(s): 4791 Problem Description 省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能间接通过公路可达即可).现得到城镇道路统计表,表中列出了任意两…

思路:求最小生成树(最小生成树就是权值之和最小的极小连通子图) ,注意将已修过的边的权值置为0: 数据结构:由于数据量小,可以用临接矩阵直接存储图 #include<stdio.h> #include<string.h> int vis[101]; int price[101]; int map[101][101]; void init(int n) { int i; memset(vis,0,sizeof(vis)); for(i = 1;i <= n;i ++) pric…

Problem Description 省政府"畅通工程"的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能间接通过公路可达即可).现得到城镇道路统计表,表中列出了任意两城镇间修建道路的费用,以及该道路是否已经修通的状态.现请你编写程序,计算出全省畅通需要的最低成本. Input 测试输入包含若干测试用例.每个测试用例的第1行给出村庄数目N ( 1< N < 100 ):随后的 N(N-1)/2 行对应村庄间道路的成本及修建状态,每行给4个正整…

首先,贴上一个很好的讲解贴: http://www.wutianqi.com/?p=3012 HDOJ 1233 还是畅通工程 http://acm.hdu.edu.cn/showproblem.php?pid=1233 裸的Prim... #include<cstdio> #define MAXN 105 #define INF 0x3f3f3f3f int map[MAXN][MAXN]; int dist[MAXN]; int vis[MAXN]; int n,a,b,x,ans,tot…

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…