最小生成树示例程序_Prim算法
May 3, 2013
Graph
输入:顶点个数n和边数m,然后是m条边的数据。u v w 分别代表两个顶点和权值。顶点从1开始记起。 输出:一次选择的各条边和最小生成树的权。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
using namespace std;
typedef long long int LL;
const int MAXN = 0x3f3f3f3f;
const int MIN = -0x3f3f3f3f;
const double eps = 1e-9;
const int dir[8][2] = {{0,1},{1,0},{0,-1},{-1,0},{-1,1},
{1,1},{1,-1},{-1,-1}};
const int MAX = 100;
int n, m, lowcost[MAX], nearvex[MAX], edge[MAX][MAX];
void prim(int u0)
{
int i, j, sum = 0;
for (i = 1; i <= n; ++i) {
lowcost[i] = edge[u0][i]; nearvex[i] = u0;
}
nearvex[u0] = -1;
for (i = 1; i < n; ++i) {
int min = MAXN, v = -1;
for (j = 1; j <= n; ++j) {
if (nearvex[j] != -1 && lowcost[j] < min) {
min = lowcost[j]; v = j;
}
}
if (v != -1) {
printf("%d %d %d\n", nearvex[v], v, lowcost[v]);
sum += lowcost[v];
nearvex[v] = -1;
for (j = 1; j <= n; ++j) {
if (nearvex[j] != -1 && lowcost[j] > edge[v][j]) {
lowcost[j] = edge[v][j]; nearvex[j] = v;
}
}
}
}
printf("MST sum is : %d\n", sum);
}
int main(void){
#ifndef ONLINE_JUDGE
freopen("prim.in", "r", stdin);
#endif
while (~scanf("%d%d", &n, &m)) {
int i, j, u, v, w;
memset(edge, 0, sizeof(edge));
for (i = 1; i <= m; ++i) {
scanf("%d%d%d", &u, &v, &w);
edge[u][v] = edge[v][u] = w;
}
for (i = 1; i <= n; ++i) {
for (j = 1; j <= n; ++j) {
if (i == j) edge[i][j] = 0;
else if (edge[i][j] == 0) edge[i][j] = MAXN;
}
}
prim(1);
}
return 0;
}
写代码的时候,还是会出现各种错误,比如for循环里面到底是不是要取到等号,一定要想清楚,还有就是细节,输入的m和n不要搞错了。