Truck History最小生成树

Truck History
原题链接https://vjudge.net/contest/352170#problem/E


题意为有 n行字母 他们之间的距离就是不同字母的个数，相当于题目将所有的地点给出，我们只需要枚举写出地图，正常计算即可。
注意Kruskal很容易记忆化超限，
Prim：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iostream>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
long long map[3005][3005];
long long dis[300005];
bool vis[300005];
long long ans;
long long n;
long long maxx;
struct node 
{
    char s[35];
}stu[39005];
void prim()
{
    ans = 0;
    long long i, j;
    for (i = 1; i <= n; i++)
    {
        dis[i] = map[1][i];
        vis[i] = false;
    }
    vis[1] = true;
    for (i = 0; i < n - 1; i++)
    {
        long long minn = INF;
        long long p = 1;
        for (j = 1; j <= n; j++)
        {
            if (!vis[j] && dis[j] < minn)
            {
                minn = dis[j];
                p = j;
            }
        }
        // cout << "***" << endl;
        //  for (j = 1; j <= n; j++)
        //  {
        //      cout << dis[j] << " ";
        //   }
        //    cout << endl;
        if (minn == INF)
        {
            ans = -1;
            return;
        }
        ans += minn;
        vis[p] = true;
        for (j = 1; j <= n; j++)
        {
            if (!vis[j] && dis[j] > map[p][j])
            {
                dis[j] = map[p][j];
            }
        }
    }
    return;
}
int main()
{
    while(~scanf("%lld",&n))
    {
        if(n==0)
        {
            break;
        }
        long long i, j;
        for (i = 1; i <= n;i++)
        {
                scanf("%s", stu[i].s);
        }
        for (i = 1; i <= n;i++)
        {
            for (j = 1; j <= n;j++)
            {
                if (i == j)
                {
                    map[i][j] = 0;
                }
                else 
                {
                    long long z,sum = 0;
                    for (z = 0; z < 7;z++)
                    {
                        if(stu[i].s[z]!=stu[j].s[z])//判断每两个字符串之间不同字母的个数。
                        {
                            sum++;
                        }
                    }
                    map[i][j] = sum;
                    map[j][i] = sum;
                }
            }
        }
        prim();
        printf("The highest possible quality is 1/%lld.\n", ans);
    }
    return 0;
}

Kruskal:

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iostream>
#include <queue>
using namespace std;
long long pre[300005];
char map[3005][20];
struct node
{
	long long u;
	long long v;
	long long w;
} stu[3000005];
long long zz;
long long cnt;
long long ans;
bool cmp(node x, node y)
{
	return x.w < y.w;
}
long long find(long long x)
{
	if (x == pre[x])
	{
		return x;
	}
	return pre[x] = find(pre[x]);
}
void join(long long x, long long y, long long w)
{
	long long ss1 = find(x);
	long long ss2 = find(y);
	if (ss2 != ss1)
	{
		pre[ss2] = ss1;
		cnt--;
		ans += w;
	}
}
int main()
{
	long long n, i, j;
	while (~scanf("%lld", &n))
	{
		if (n == 0)
		{
			break;
		}
		ans = 0;
		cnt = n;
		long long num = 1;
		for (i = 0; i <= n; i++)
		{
			pre[i] = i;
		}
		for (i = 1; i <= n; i++)
		{
			scanf("%s", map[i]);
		}
//		for (i = 1; i <= n; i++)
//		{
//			cout << map[i] << endl;
//		}
		long long k = 0;
		for (i = 1; i < n; i++)
		{
			for (j = i + 1; j <= n; j++)
			{
				stu[k].u = i;
				stu[k].v = j;
				long long sum = 0;
			//	cout << map[i] << "  " << map[j] << endl;
				for (zz = 0; zz < 7; zz++)
				{
					if (map[i][zz] != map[j][zz])
					{
						sum++;
					}
				}
				stu[k].w = sum;//这里注意直接读取结果，不要装进二维数组中再读取，会记忆化超限。
			//	cout << "sum=" << sum << endl;
				k++;
			}
		}
		sort(stu, stu + k, cmp);
		//for (i = 0; i < k; i++)
		//{
		//	cout << stu[i].u << " " << stu[i].v << " " << stu[i].w << endl;
		//}
		for (i = 0; i < k; i++)
		{
			join(stu[i].u, stu[i].v, stu[i].w);
			if (cnt == 1)
			{
				break;
			}
		}
		printf("The highest possible quality is 1/%lld.\n", ans);
	}
	return 0;

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来源： https://blog.csdn.net/yeyuluo/article/details/104064000