#include <stdio.h>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int father[510],co[510][510];
struct node{
int u,v,w;
}qwe[250000];
bool cmp(node a,node b)
{
return a.w<b.w;
}
int find(int x)
{
if(x==father[x]) return x;
else return father[x]=find(father[x]);
}
int unite(int x,int y)
{
x=find(x);
y=find(y);
if(x!=y)
{
father[x]=y;
}
}
int kual(int l)
{
int ans=0;
for(int i=0;i<l;i++)
{
int x=find(qwe[i].u);
int y=find(qwe[i].v);
if(x!=y)
{
father[x]=y;
ans+=qwe[i].w;
}
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
father[i]=i;
}
int k,l=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&k);
qwe[l].u=i;
qwe[l].v=j;
qwe[l].w=k;
l++;
}
}
// sort(qwe,qwe+l,cmp);
int q,c,d,sum=0;
scanf("%d",&q);
for(int i=0;i<q;i++)
{
scanf("%d%d",&c,&d);//首先处理已经连接的点
for(int j=0;j<l;j++)
{
if((qwe[i].u==c&&qwe[i].v==d)||(qwe[i].v==d&&qwe[i].u==c))
{
qwe[i].w=0;
break;
}
}
unite(c,d);
}
sort(qwe,qwe+l,cmp);
printf("%d\n",kual(l));
}
return 0;
}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.
题解：就是先给出已经连接的点，求出让所有点都联通的至少还需要多少
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
#include <stdio.h>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int father[510],co[510][510];
struct node{
int u,v,w;
}qwe[250000];
bool cmp(node a,node b)
{
return a.w<b.w;
}
int find(int x)
{
if(x==father[x]) return x;
else return father[x]=find(father[x]);
}
int unite(int x,int y)
{
x=find(x);
y=find(y);
if(x!=y)
{
father[x]=y;
}
}
int kual(int l)
{
int ans=0;
for(int i=0;i<l;i++)
{
int x=find(qwe[i].u);
int y=find(qwe[i].v);
if(x!=y)
{
father[x]=y;
ans+=qwe[i].w;
}
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
father[i]=i;
}
int k,l=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&k);
qwe[l].u=i;
qwe[l].v=j;
qwe[l].w=k;
l++;
}
}
// sort(qwe,qwe+l,cmp);
int q,c,d,sum=0;
scanf("%d",&q);
for(int i=0;i<q;i++)
{
scanf("%d%d",&c,&d);//首先处理已经连接的点
for(int j=0;j<l;j++)
{
if((qwe[i].u==c&&qwe[i].v==d)||(qwe[i].v==d&&qwe[i].u==c))
{
qwe[i].w=0;
break;
}
}
unite(c,d);
}
sort(qwe,qwe+l,cmp);
printf("%d\n",kual(l));
}
return 0;
}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