POJ 1258 Agri-Net （经典MST）

链接：http://poj.org/problem?id=1258

Description

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course. 
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms. 
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm. 
The distance between any two farms will not exceed 100,000. 

Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

Sample Input

40 4 9 214 0 8 179 8 0 1621 17 16 0

Sample Output

28

解题思路： 最简单的最小生成树的题目；MST-Kruscal

代码如下：(Kruscal)

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <string>#include <iomanip>#include <cassert>#include <algorithm>#define LC(x) (x<<1)#define RC(x) (LC(x)+1)#define PI (acos(-1))#define EPS 1e-8#define MAXN 111111#define MAXM 222222#define LL long long#define ULL unsigned long long#define INF 0x7fffffff#define pb push_back#define mp make_pair#define lowbit(x) (x&(-x))#define RST(N)memset(N, 0, sizeof(N))using namespace std;struct edge {    int v, u, cost;}e[111*111];bool cmp_e(const edge &a, const edge &b){    return a.cost < b.cost;}int ecnt, fa[111], n, v;void addedge(int u, int v, int cost) {    e[++ ecnt].v = v, e[ecnt].u = u, e[ecnt].cost = cost;}int getf(int x){    return fa[x] == x ? x : fa[x] = getf(fa[x]);}int kruscal() {    int sum = 0;    for(int i = 1; i <= n; i ++) fa[i] = i;    sort(e + 1, e + 1 + ecnt, cmp_e);    for(int i = 1; i <= ecnt; i ++) {        int u = e[i].u, v = e[i].v;        u = getf(u), v = getf(v);        if(u == v) continue;        sum += e[i].cost;        fa[u] = v;    }    return sum;}int main(){    while(scanf("%d", &n) == 1) {        ecnt = 0;        for(int i = 1; i <= n; i ++) {            for(int j = 1; j <= n; j ++) {                scanf("%d", &v);                if(v != 0) addedge(i, j, v);            }        }        printf("%d\n", kruscal());    }    return 0;}

Prim:

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#define MAXN 105#define RST(N)memset(N, 0, sizeof(N))using namespace std;int a[MAXN][MAXN], n;int flag[MAXN], tot;  //flag[i]用来标记节点i是否被覆盖void Init(){    for(int i=1; i<=n; i++) {        for(int j=1; j<=n; j++) {            scanf("%d", &a[i][j]);        }    }    RST(flag), tot = 0;}int main(){    while(~scanf("%d", &n)) {        Init();        flag[1] = 1;       //选取第一个点        for(int k=1; k<n; k++) {       //循环n-1次            int min = -1, min_i;            for(int i=1; i<=n; i++) {       //选取下一个最小权值的节点                if(flag[i] == 0 && (min == -1 || a[1][i] < min)) {                    min = a[1][i]; min_i = i;                }            }            flag[min_i]=1;    //覆盖节点            for(int i=1; i<=n; i++) {        //更新未覆盖节点的距离                if(flag[i] == 0 && a[1][i] > a[min_i][i]) {                    a[1][i] = a[min_i][i];                }            }            tot += a[1][min_i];   //加上权值        }        printf("%d\n", tot);    }    return 0;}