Truck History(kruskal+prime)
Truck History
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 59 Accepted Submission(s) : 21
Today, ACM is rich enough to pay historians to study its history. One thing historians tried to find out is so called derivation plan -- i.e. how the truck types were derived. They defined the distance of truck types as the number of positions with different letters in truck type codes. They also assumed that each truck type was derived from exactly one other truck type (except for the first truck type which was not derived from any other type). The quality of a derivation plan was then defined as
1/Σ(to,td)d(to,td)
where the sum goes over all pairs of types in the derivation plan such that to is the original type and td the type derived from it and d(to,td) is the distance of the types.
Since historians failed, you are to write a program to help them. Given the codes of truck types, your program should find the highest possible quality of a derivation plan.
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include <algorithm>
using namespace std;
const int MAXN=; int pre[MAXN],anser,num,N; struct Node{
int s,e,dis;
}; Node dt[MAXN*MAXN];
char str[MAXN][]; int gd(char *a,char *b){
int t=;
for(int i=;a[i];i++){
if(a[i]!=b[i])t++;
}
return t;
} int find(int x){
//return pre[x]= x==pre[x]?x:find(pre[x]);
int r=x;
while(r!=pre[r])r=pre[r];
int i=x,j;
while(i!=r)j=pre[i],pre[i]=r,i=j;
return r;
}
/*void merge(Node a){
int f1,f2;
if(num==N)return;
if(pre[a.s]==-1)pre[a.s]=a.s;
if(pre[a.e]==-1)pre[a.e]=a.e;
f1=find(a.s);f2=find(a.e);
if(f1!=f2){num++;
pre[f1]=f2;
anser+=a.dis;
}
}*/
void initial(){
memset(pre,-,sizeof(pre));
//memset(dt,0,sizeof(dt));
//memset(str,0,sizeof(str));
anser=;
} //int cmp(const void *a,const void *b){
// if((*(Node *)a).dis<(*(Node *)b).dis)return -1;
// else return 1;
//} int cmp(Node a, Node b){
return a.dis < b.dis;
} int main(){
while(scanf("%d",&N),N){
num=;
initial();
for(int i=;i<N;i++){
scanf("%s",str[i]);
}
int k=;
for(int i=;i<N - ;i++){
for(int j=i+;j<N;j++){
dt[k].s=i;
dt[k].e=j;
dt[k].dis=gd(str[i],str[j]);
// printf("k==%d %d %d %d\n ",k,i,j,dt[k].dis);
k++;
}
}
sort(dt, dt + k, cmp);
//qsort(dt,k,sizeof(dt[0]),cmp);
int f1,f2;
for(int i=;i<k;i++){
if(pre[dt[i].s]==-)pre[dt[i].s]=dt[i].s;
if(pre[dt[i].e]==-)pre[dt[i].e]=dt[i].e;
f1=find(dt[i].s);
f2=find(dt[i].e);
if(f1!=f2){
num++;
pre[f1]=f2;
anser+=dt[i].dis;
}
if(num==N)break;
//merge(dt[i]);
}
printf("The highest possible quality is 1/%d.\n",anser);
}
return ;
}
prime代码:
#include<stdio.h>
#include<string.h>
const int INF=0x3f3f3f3f;
const int MAXN=;
int map[MAXN][MAXN],low[MAXN];
int vis[MAXN];
int N,ans;
char str[MAXN][];
void prime(){
memset(vis,,sizeof(vis));
int temp,k,flot=;
vis[]=;
for(int i=;i<=N;i++)low[i]=map[][i];
for(int i=;i<=N;i++){
temp=INF;
for(int j=;j<=N;j++)
if(!vis[j]&&temp>low[j])
temp=low[k=j];
if(temp==INF){
printf("The highest possible quality is 1/%d.\n",ans);
break;
}
ans+=temp;
flot++;
vis[k]=;
for(int j=;j<=N;j++)
if(!vis[j]&&map[k][j]<low[j])
low[j]=map[k][j];
}
}
int fd(char *a,char *b){
int t=;
for(int i=;a[i];i++){
if(a[i]!=b[i])t++;
}
return t;
}
void initial(){
memset(map,INF,sizeof(map));
ans=;
}
int main(){int s;
while(~scanf("%d",&N),N){
initial();
for(int i=;i<=N;i++)scanf("%s",str[i]);
for(int i=;i<=N;i++){
for(int j=i+;j<=N;j++){
s=fd(str[i],str[j]);
if(s<map[i][j])map[i][j]=map[j][i]=s;
}
}
prime();
}
return ;
}
Truck History(kruskal+prime)的更多相关文章
- poj 1789 Truck History(kruskal算法)
主题链接:http://poj.org/problem?id=1789 思维:一个一个点,每两行之间不懂得字符个数就看做是权值.然后用kruskal算法计算出最小生成树 我写了两个代码一个是用优先队列 ...
- Truck History(卡车历史)
Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 26547 Accepted: 10300 Description Adv ...
- Truck History(poj 1789)
Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for v ...
- POJ 1789 Truck History(Prim+邻接矩阵)
( ̄▽ ̄)" #include<iostream> #include<cstdio> #include<cstring> #include<algo ...
- 最小生成树---普里姆算法(Prim算法)和克鲁斯卡尔算法(Kruskal算法)
普里姆算法(Prim算法) #include<bits/stdc++.h> using namespace std; #define MAXVEX 100 #define INF 6553 ...
- Truck History(prime)
Truck History Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 31871 Accepted: 12427 D ...
- POJ 1789 Truck History (最小生成树)
Truck History 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/E Description Advanced Carg ...
- POJ 1789:Truck History(prim&&最小生成树)
id=1789">Truck History Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17610 ...
- Truck History(最小生成树)
poj——Truck History Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 27703 Accepted: 10 ...
随机推荐
- Java开发者工具
From:http://www.csdn.net/article/2015-03-26/2824317 1. Notepad++ Notepad++是用于编辑xml.脚本以及记笔记的最佳工具.这个工具 ...
- UESTC_方老师的分身 II CDOJ 915
方老师的分身 II Time Limit: 10000/5000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others) Submi ...
- POJ1270 Following Orders (拓扑排序)
Following Orders Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4254 Accepted: 1709 ...
- Multiple outputs from T4 made easy – revisited » DamienG
Multiple outputs from T4 made easy – revisited » DamienG Multiple outputs from T4 made easy – revisi ...
- 修改MySQL 5.5的max_allowed_packet属性的方法
今天在部署一个实验系统的时候,报出下面这个错: Your 'max_allowed_packet' variable is set to less than 16777216 Byte (16MB). ...
- OpenGL进阶(十一) - GLSL4.x中的数据传递
in out 对于 vertex shader,每个顶点都会包含一次,它的主要工作时处理关于定点的数据,然后把结果传递到管线的下个阶段. 以前版本的GLSL,数据会通过一些内建变量,比如gl_Vert ...
- 牵一发动全身【Nhibernate基本映射】
用牵一发动全身来形容Nhibernate的映射,一点都不夸张.小小的属性的修改,决定了整个Nhibernate的执行动态.以下让我们来详细了解一下,通过回想我们在上篇文章中用到的配置文件,做一个对xm ...
- Linux-0.11内核源代码分析系列:内存管理get_free_page()函数分析
Linux-0.11内存管理模块是源码中比較难以理解的部分,如今把笔者个人的理解发表 先发Linux-0.11内核内存管理get_free_page()函数分析 有时间再写其它函数或者文件的:) /* ...
- 查看Linux操作系统版本
1.查看内核版本命令: [root@server1 Desktop]# cat /proc/version Linux version 2.6.32-358.el6.x86_64 (mockbui ...
- JavaScript脚本语言的正则校验法
正则校验法有很多种类型,有些可能会比较复杂难记,我这里罗列了大家常用的几种方法,方便查询. //校验是否全由数字组成 function isShuZi(s) { var patrn=/^[0-9]{1 ...