POJ 2075 －－最小生成树算法

Tangled(纠缠不清) in Cables（电缆）

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 959		Accepted: 461

Description

You are the owner of SmallCableCo and have purchased（购买） the franchise（专营权） 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（仓库） you bought. Among your finds is a single spool(阀芯） of cable and a lot of connectors（连接器）. 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（实数）.
  
• 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（标点符号）.
  
• Next line: M, number of paths between houses
  
• next M lines in the form（形式）

< 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（样品） 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
代码如下：

 

#include <stdio.h>
#include <string.h>

#define MAX 32767
int main(int argc, char **argv)
{
        double cables, value, min, result = 0.0;
        int i, j, k, names, paths, row, col;
        char buf1[25], buf2[25];

        scanf("%lf", &cables);
        scanf("%d", &names);
        int nodes[names], closest[names];
        char name[names][25];
        double lowcost[names], path[names][names];

        memset(nodes, 0, sizeof(nodes));
        memset(lowcost, 0, sizeof(lowcost));
        memset(closest, 0, sizeof(closest));
        memset(name, 0, sizeof(name));

        for(i = 0; i < names; i++){
                for(j = 0; j < names;j++){
                        path[i][j] = MAX;
                }
        }

        for(i = 0; i < names; i++){
                scanf("%s", name[i]);
        }

        scanf("%d", &paths);
        for(i = 0; i < paths; i++){
                row = col = 0;
                scanf("%s %s %lf", buf1, buf2, &value);
                for(j = 0; j < names; j++){
                        if(strcmp(buf1, name[j]) == 0){
                                row = j;
                        }
                        else if(strcmp(buf2, name[j]) == 0){
                                col = j;
                        }
                }
                path[row][col] = path[col][row] = value;
        }

        nodes[0] = 1;
        for(i = 1; i < names; i++){
                lowcost[i] = path[0][i];
                closest[i] = 0;
        }

        for(i = 1; i < names; i++){
                min = MAX;
                j = 0;
                for(k = 1; k < names; k++){
                        if(!nodes[k] && lowcost[k] && lowcost[k] < min){
                                min = lowcost[k];
                                j = k;
                        }
                }
                nodes[j] = 1;
                result += min;
                for(k = 1; k < names; k++){
                        if(!nodes[k] && path[j][k] && path[j][k] < lowcost[k]){
                                lowcost[k] = path[j][k];
                                closest[k] = j;
                        }
                }
        }
        if(result > cables){
                printf("Not enough cable\n");
        }
        else{
                printf("Need %.1lf miles of cable\n", result);
        }

        return 0;
}