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數據結構與算法問題北大oj 2075（最小生成樹）

編輯：C++入門知識

數據結構與算法問題北大oj 2075（最小生成樹）

Tangled in Cables Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5943 Accepted: 2342

Description

You are the owner of SmallCableCo and have purchased the franchise rights for a small town. Unfortunately, you lack enough funds to start your business properly and are relying on parts you have found in an old warehouse you bought. Among your finds is a single spool of cable and a lot of connectors. You want to figure out whether you have enough cable to connect every house in town. You have a map of town with the distances for all the paths you may use to run your cable between the houses. You want to calculate the shortest length of cable you must have to connect all of the houses together.

Input

Only one town will be given in an input.
The first line gives the length of cable on the spool as a real number.
The second line contains the number of houses, N
The next N lines give the name of each house's owner. Each name consists of up to 20 characters {a–z,A–Z,0–9} and contains no whitespace or punctuation.
Next line: M, number of paths between houses
next M lines in the form
< house name A > < house name B > < distance >
Where the two house names match two different names in the list above and the distance is a positive real number. There will not be two paths between the same pair of houses.

Output

The output will consist of a single line. If there is not enough cable to connect all of the houses in the town, output
Not enough cable
If there is enough cable, then output
Need < X > miles of cable
Print X to the nearest tenth of a mile (0.1).

Sample Input

100.0
4
Jones
Smiths
Howards
Wangs
5
Jones Smiths 2.0
Jones Howards 4.2
Jones Wangs 6.7
Howards Wangs 4.0
Smiths Wangs 10.0

Sample Output

Need 10.2 miles of cable.

因為需要輸入字符，我用了STL中的map映射，將地址名字符串映射為1,2,3....n.

代碼：

#include 
#include 
#include 
using namespace std;
const int IN = 65535;
typedef struct graph
{
	string vex[200];
	double adj[200][200];
	int numvex, numedge;
}graph;

double num;
void create(graph *g)
{
	int i, j,x1,x2;
	string s1, s2;
	double w;//權值
	map s;
	cin >> num;
	cin >> g->numvex;
	for (i = 0; i < g->numvex; i++)
	{
		cin >> g->vex[i];
		s[g->vex[i]] = i;
	}
	cin >> g->numedge;
	for (i = 0; i < g->numedge; i++)
		for (j = 0; j < g->numedge; j++)
			g->adj[i][j] = IN;

	for (j = 0; j < g->numedge; j++)
	{
		cin >> s1 >> s2 >> w;
		x1 = s[s1];
		x2 = s[s2];
		g->adj[x1][x2] = w;
		g->adj[x2][x1] = g->adj[x1][x2];
	}

}

void prim(graph *g)
{
	int i, j, k;
	double bowcost[200];
	double min, num1 = 0;
	int adjvex[200];
	bowcost[0] = 0;
	adjvex[0] = 0;
	for (i = 1; i < g->numvex; i++)
	{
		adjvex[i] = 0;
		bowcost[i] = g->adj[0][i];
	}
	for (i = 1; i < g->numvex; i++)
	{
		min = IN;
		j = 1;
		while (j < g->numvex)
		{
			if (bowcost[j] != 0 && bowcost[j] < min)
			{
				min = bowcost[j];
				k = j;
			}
			j++;
			
		}
		num1 += min; 
		
		bowcost[k] = 0;
		for (j = 1; j < g->numvex; j++)
		{
			if (g->adj[k][j] < bowcost[j]&&bowcost[j] != 0)
			{
				bowcost[j] = g->adj[k][j];
				adjvex[j] = k;
			}
		}

	}
	cout << "Need" << num << "miles of cable" << endl;
}

int main()
{
	graph *g = new graph;
	create(g);
	prim(g);
}