HDOJ 1102 Constructing Roads(最小生成树)
Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6455 Accepted Submission(s): 2377
Problem Description
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village
C such that there is a road between A and C, and C and B are connected.
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village
i and village j.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
Sample Input
3 0 990 692 990 0 179 692 179 0 1 1 2
Sample Output
179
Source
这道题的思路就是求最小生成树。详见AC code 的注释
先贴一个WA的代码，这个代码的bug让我查了两天，最后才茅塞顿开。。。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#define max 0x7fffffff
using namespace std;
struct edge
{
int v1;
int v2;
int w;
}e[6000];
int cmp(const void *a,const void *b)
{
struct edge *aa=(struct edge *)a;
struct edge *bb=(struct edge *)b;
if(aa->w != bb->w)
return aa->w - bb->w;
else
return aa->v1 - bb->v1;
}
int main()
{
int n,q,a,b,map[101][101],vis[101],i,j,k,min;
while(scanf("%d",&n)!=EOF)
{
min=0;
for(i=1;i<=n;i++) vis[i]=i;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&map[i][j]);
}
map[i][i]=max;
}
scanf("%d",&q);
for(i=1;i<=q;i++)
{
scanf("%d%d",&a,&b);
map[a][b]=map[b][a]=0;
vis[b]=vis[a];//连通的两点标记相同，修改数值大的顶点使之与小的顶点相同，bug就在这里
}
for(i=1,k=1;i<=n;i++)
{
for(j=1;j<=i;j++)
{
e[k].v1=i;
e[k].v2=j;
e[k].w=map[i][j];
k++;
}
}
qsort(&e[1],k-1,sizeof(e[1]),cmp);
//j记录生成树中的边数，边数最终要等于n-1
for(i=1,j=q;j<n-1;i++)
{
if(vis[e[i].v1] != vis[e[i].v2])
{
int m=vis[e[i].v1]>vis[e[i].v2] ? vis[e[i].v2] : vis[e[i].v1];
int M=vis[e[i].v1]>vis[e[i].v2] ? vis[e[i].v1] : vis[e[i].v2];
for(int ii=1;ii<=n;ii++)
{
if(vis[ii]==M)
vis[ii]=m;
}
min+=e[i].w;
j++;
}
}
printf("%d\n",min);
}
return 0;
}
出现错误的一个例子：已经修好的路有两条：2,4和1,2.当输入2,4时，修改4的标记使之与2相同，当输入1,2时修改2的标记使之与1相同，注意这时没有同时修改4，这样一来从标记上看4与2不连通，后面就出错。我原来一直以为一定是1,2；2,4这样输入，所以一直没发现错误。
AC code（15ms）
#include <iostream>
#include <cstdio>
#include <cstdlib>
#define max 0x7fffffff
using namespace std;
struct edge
{
int v1;
int v2;
int w;
}e[6000];//储存边
int cmp(const void *a,const void *b)
{
struct edge *aa=(struct edge *)a;
struct edge *bb=(struct edge *)b;
if(aa->w != bb->w)
return aa->w - bb->w;
else
return aa->v1 - bb->v1;//当权值相同时按顶点排序
}
int main()
{
int n,q,a,b,map[101][101],vis[101],i,j,k,min;//map记录邻接矩阵，vis为顶点设置标志
while(scanf("%d",&n)!=EOF)
{
min=0;
for(i=1;i<=n;i++) vis[i]=i;//初始化vis，使各个顶点开始时各自属于独立的集合
//输入邻接矩阵
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&map[i][j]);
}
map[i][i]=max;//自己到自己的路径修改为max
}
//输入已修好的路径
scanf("%d",&q);
for(i=1;i<=q;i++)
{
scanf("%d%d",&a,&b);
map[a][b]=map[b][a]=0;
//vis[b]=vis[a];
}
//将边的信息存入结构体e，注意只需存下三角形
for(i=1,k=1;i<=n;i++)
{
for(j=1;j<=i;j++)
{
e[k].v1=i;
e[k].v2=j;
e[k].w=map[i][j];
k++;
}
}
//快排，因为从e[1]开始储存，故从e[1]开始排序
qsort(&e[1],k-1,sizeof(e[1]),cmp);
//生成树
for(i=1,j=1;j<n;i++)
{
//两个顶点属于不同集合时表明边尚未加入树中
if(vis[e[i].v1] != vis[e[i].v2])
{
//修改数值大的顶点的标志及其所有祖先的标志，将他们纳入树中
int m=vis[e[i].v1]>vis[e[i].v2] ? vis[e[i].v2] : vis[e[i].v1];
int M=vis[e[i].v1]>vis[e[i].v2] ? vis[e[i].v1] : vis[e[i].v2];
for(int ii=1;ii<=n;ii++)
{
if(vis[ii]==M)
vis[ii]=m;
}
min+=e[i].w;
j++;
}
}
printf("%d\n",min);
}
return 0;
}
