prim算法 求最小生成树 1075

生成树 贪心

求出给定无向带权图的最小生成树。图的定点为字符型，权值为不超过100的整形。在提示中已经给出了部分代码，你只需要完善Prim算法即可。

Description

第一行为图的顶点个数n
第二行为图的边的条数e
接着e行为依附于一条边的两个顶点和边上的权值

Input

最小生成树中的边。

Output

1

2

3

4

5

6

7

8

9

10

11

12

13

6

10

ABCDEF

A B 6

A C 1

A D 5

B C 5

C D 5

B E 3

E C 6

C F 4

F D 2

E F 6

Sample Input

1

(A,C)(C,F)(F,D)(C,B)(B,E)

#include<stdio.h>#include<string>#include<iostream>#include<algorithm>#include<stack>#include<queue>#include<string.h>using namespace std;#define mem(x,y) memset(x,y,sizeof(x))int maps[105][105];char dian[1000];void pri(int n, int e){int disx[1000];int close[1000];for (int i = 0; i < e; i++){disx[i] = maps[0][i];close[i] = 0;}for (int i = 1; i < n; i++){int min = 0xfff, k;for (int j = 0; j < n; j++){if (disx[j] != 0 && disx[j] < min){k = j;min = disx[j];}}cout << '(' << dian[close[k]] << ',' << dian[k] << ')';disx[k] = 0;for (int j = 0; j < n; j++){if (maps[k][j] != 0 && maps[k][j] < disx[j]){disx[j] = maps[k][j];close[j] = k;}}}}int main(){int n;int e;while (cin >> n){cin >> e;scanf("%s", dian);for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){maps[i][j] = i == j ? 0 : 0xfff;}}for (int i = 0; i < e; i++){char from, to;int dis;cin >> from >> to >> dis;maps[from - 'A'][to - 'A'] = dis;maps[to - 'A'][from - 'A'] = dis;}pri(n, e);}return 0;}

//by tp