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  • POJ 2421 Constructing Roads

    Language:
    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
    ",sum);
        return 0;
    }
    View Code


    作者:乌鸦坐飞机
    出处:http://www.cnblogs.com/whistle13326/
    新的风暴已经出现 怎么能够停止不前 穿越时空 竭尽全力 我会来到你身边 微笑面对危险 梦想成真不会遥远 鼓起勇气 坚定向前 奇迹一定会出现

     
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  • 原文地址:https://www.cnblogs.com/whistle13326/p/6364230.html
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