poj 2421 Constructing Roads（最小生成树）

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20139		Accepted: 8443

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

需要建路 给出点与点之间的距离 现在有些点之间已经有路了 求出现在建路的最小距离
如果已经建路了的 则用flag标记 且这条边的权值为0

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 1010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char ch;
    int a = 0;
    while((ch = getchar()) == ' ' | ch == '\n');
    a += ch - '0';
    while((ch = getchar()) != ' ' && ch != '\n')
    {
        a *= 10;
        a += ch - '0';
    }
    return a;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}
int mp[MAXN][MAXN];
int flag[MAXN][MAXN];
int F[MAXN];
struct Edge
{
    int u,v,w;
}edge[MAXM];
int tol;
void addedge(int u,int v,int w)
{
    edge[tol].u=u;
    edge[tol].v=v;
    edge[tol++].w=w;
}
bool cmp(Edge a,Edge b)
{
    return a.w<b.w;
}
int find(int x)
{
    if(F[x]==-1)  return x;
    return F[x]=find(F[x]);
}

int Kruskal(int n)
{
    MEM(F,-1);
    sort(edge,edge+tol,cmp);
    int cnt=0;
    int ans=0;
    for(int i=0;i<tol;i++)
    {
        int u=edge[i].u;
        int v=edge[i].v;
        int w=edge[i].w;
        int t1=find(u);
        int t2=find(v);
        if(t1!=t2)
        {
            ans+=w;
            F[t1]=t2;
            cnt++;
        }
        if(cnt==n-1) break;
    }
    if(cnt<n-1) return -1;
    return ans;
}

int main()
{
//    fread;
    int n,m;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&mp[i][j]);
            }
        }
        scanf("%d",&m);
        MEM(flag,0);
        while(m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            flag[u][v]=flag[v][u]=1;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(flag[i][j])
                    addedge(i,j,0);
                else addedge(i,j,mp[i][j]);
            }
        }
        int ans=Kruskal(n);
        printf("%d\n",ans);
    }
    return 0;
}