POJ-2421Constructing Roads,又是最小產生樹，和第八屆河南省賽的引水工程驚人的相似，並查集與最小產生樹的靈活與能用，水過~~~

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Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K

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

30 990 692990 0 179692 179 011 2

Sample Output

179

Source

PKU Monthly,kicc

   找的最小產生樹專題裡的一個題，看到這題卻發現和河南省第八屆省賽的“引水工程”驚人的相似，反正那道題也沒做出來，本打算先A了那道題再做這道題，可是做了那題之後又發現這道題還更簡單，於是調調代碼直接A了，其實我寧願相信是後台測試資料水的；

    來說說思路吧:我們發現輸入是以矩陣的模式，如果全部存在結構體中無疑是浪費記憶體，我們發現這個矩陣中很多相同的數，即座標相對應的點（如a[i][j]=a[j][i]）的值是一樣的,所以只需要把矩陣中的一半的數存在結構體中就可以了，然後就是最簡單的最小產生樹了，至於有Q條已經聯通好的路我們只需在輸入得時候利用並查集將輸入的資料的根節點合并即可；可能這裡表達有點不清楚，來看代碼就馬上明白了；

       AC代碼：說白了，這題就是並查集與最小產生樹的靈活運用；

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<algorithm>using namespace std;const int N=5100;//資料範圍是100，平方是10000，但Q<=10000/2,開5050就可以過了；int n,m,k,f[100];struct node{    int u,v,w;} a[N];int cmp(node a,node b){    return a.w<b.w;}int find(int x){    return f[x]==-1?x:x=find(f[x]);}int ks(int n){    int ans=0,cot=0;    sort(a+1,a+1+k,cmp);    for(int i=1; i<=k; i++)    {        int u=find(a[i].u);        int v=find(a[i].v);        if(u!=v)        {            ans+=a[i].w;            f[u]=v;            cot++;        }        if(cot==n-1)            break;    }    return ans;}int main(){    int x;    scanf("%d",&n);    k=0;    memset(a,0,sizeof(a));    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)        {            scanf("%d",&x);            if(j>i)//把一半的資料存在結構體中；                a[++k].u=i,a[k].v=j,a[k].w=x;        }    memset(f,-1,sizeof(f));    scanf("%d",&m);    int  x1,y1;    while(m--)    {        scanf("%d%d",&x1,&y1);        int xx=find(x1);        int yy=find(y1);        if(xx!=yy)//根節點合并->並查集；        {            f[xx]=yy;            n--;        }    }    printf("%d\n",ks(n));    return 0;}

POJ-2421Constructing Roads,又是最小產生樹，和第八屆河南省賽的引水工程驚人的相似，並查集與最小產生樹的靈活與能用，水過~~~