POJ 2421 Constructing Roads

Posted 2020-08-29 心之所向 素履以往

Language: Default Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 23333   Accepted: 10012 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

思路：裸地MST，只是他给出你有几条道路已经建好，只需要将建好的道路赋值为0即可

#include<cstdio>
#include<iostream>
#include<algorithm>
#define MAXM 2000001
#define MAXN 201
using namespace std;
struct node {
    int x,y;
    int val;
};
node a[MAXM];
int map[MAXN][MAXN];
int fa[MAXN],n,m,cnt,tot,sum;
inline void read(int&x) {
    x=0;int f=1;char c=getchar();
    while(c>\'9\'||c<\'0\') {if(c==\'-\') f=-1;c=getchar();}
    while(c>=\'0\'&&c<=\'9\') {x=(x<<1)+(x<<3)+c-48;c=getchar();}
    x=x*f;
}
inline bool cmp(node x,node y) {
    return x.val<y.val;
}
inline int find(int x) {
    if(x==fa[x]) return x;
    else return fa[x]=find(fa[x]);
}
inline void kurskal() {
    sort(a+1,a+cnt+1,cmp);
    for(int i=1;i<=cnt;i++) {
        int xx=find(a[i].x);
        int yy=find(a[i].y);
        if(xx!=yy) {
            fa[xx]=yy;
            tot++;
            sum+=a[i].val;
        }
        if(tot==n-1) return;
    }
    
}
int main() {
    read(n);
    for(int i=1;i<=n;i++) fa[i]=i;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++) {
          read(map[i][j]);
          if(i==j) map[i][j]=1e8;
      }
    read(m);
    int x,y;
    for(int i=1;i<=m;i++) {
        read(x);read(y);
        map[x][y]=0;
        map[y][x]=0;
    }
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++) {
          a[++cnt].x=i;
          a[cnt].y=j;
          a[cnt].val=map[i][j];
      }
    kurskal();
    printf("%d\\n",sum);
    return 0;
}

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