HDOJ 1102 Constructing Roads

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18371 Accepted Submission(s): 7021



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

3
0 990 692
990 0 179
692 179 0
1
1 2

[align=left]Sample Output[/align]

179

[align=left]Source[/align]
kicc

[align=left]Recommend[/align]
Eddy

英文题，依旧是看了半天才理解题意，总觉得自己的阅读能力又上升了= =……

嗯，知识点是最小生成树。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int per[101];
int Trush[101*101];
struct Dis
{
int start,end,dis,exist;
}di[101*101];
bool cmp(Dis a,Dis b)
{
return a.dis<b.dis;
}
int Find(int r)
{
while(r!=per[r])
r=per[r];
return r;
}
int Judge(int x,int y)
{
int fx=Find(x);
int fy=Find(y);
if(fx!=fy)
{
per[fx]=fy;
return 1;
}
return 0;
}
int main()
{
int n,q,i,j,k,h,s,e;
while(~scanf("%d",&n))
{
memset(di,0,sizeof(di));
for(i=1;i<=n;i++)
per[i]=i;
for(i=1,k=0,h=0;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(i==j||i>j)
{
scanf("%d",&Trush[h]);
h++;
continue;
}
scanf("%d",&di[k].dis);
di[k].start=i;di[k].end=j;
k++;
}
}
scanf("%d",&q);
int sum=0;
while(q--)
{
scanf("%d%d",&s,&e);
for(i=0;i<k;i++)
{
if((s==di[i].start&&e==di[i].end)||(e==di[i].start&&s==di[i].end))
{
di[i].exist=1;
di[i].start=Find(di[i].start);
di[i].end=Find(di[i].end);
per[di[i].end]=di[i].start;
break;
}
}
}
sort(di,di+k,cmp);
for(i=0;i<k;i++)
{
if(di[i].exist)
continue;
if(Judge(di[i].start,di[i].end))
{
sum+=di[i].dis;
}
}
printf("%d\n",sum);
}
return 0;
}