HDOJ 1102 Constructing Roads(最小生成ツリー)
6775 ワード
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.
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.
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
エラーが発生した例:修正された道は2,4と1,2である.2,4を入力すると,修正4のマークを2と同じにし,1,2を入力すると2のマークを1と同じにするように修正し,注意すると4を同時に修正しなかったので,マークから見ると4と2がつながっておらず,裏面がエラーとなる.きっと1,2だと思っていた.2,4はこのように入力したので、ずっと間違いに気づかなかった.
AC code(15ms)
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.
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.
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
kicc
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
#include
#include
#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];// ， ，bug
}
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 ， n-1
for(i=1,j=q;jvis[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
",min);
}
return 0;
}
エラーが発生した例:修正された道は2,4と1,2である.2,4を入力すると,修正4のマークを2と同じにし,1,2を入力すると2のマークを1と同じにするように修正し,注意すると4を同時に修正しなかったので,マークから見ると4と2がつながっておらず,裏面がエラーとなる.きっと1,2だと思っていた.2,4はこのように入力したので、ずっと間違いに気づかなかった.
AC code(15ms)
#include
#include
#include
#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;//map ，vis
while(scanf("%d",&n)!=EOF)
{
min=0;
for(i=1;i<=n;i++) vis[i]=i;// vis，
//
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&map[i][j]);
}
map[i][i]=max;// 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];
}
// e，
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++;
}
}
// ， e[1] ， e[1]
qsort(&e[1],k-1,sizeof(e[1]),cmp);
//
for(i=1,j=1;jvis[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
",min);
}
return 0;
}