CodeForces 436C (Dungeons and Candies)

最新推荐文章于 2020-02-28 08:55:27 发布

j_d_m_y

最新推荐文章于 2020-02-28 08:55:27 发布

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文章标签：最小生成树 prim算法

本文链接：https://blog.youkuaiyun.com/j_d_m_y/article/details/47422269

本文介绍了一种使用C++实现的最小生成树算法，并详细展示了从输入数据到输出最小生成树的具体过程。该算法首先读取图形数据，然后计算各节点间的距离，最后通过不断寻找最短边来构造最小生成树。

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最小生成树，一次ac(兴奋ing)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct edge
{
	int x, y, w;
}e[1001];
const int N = 1e5 + 5,MAX=99999999;
int n, m, k, w;
char s[1001][121]; int dis[1001][1001], low[1001]; edge ee[1001];
int main() {
	scanf("%d%d%d%d", &n, &m, &k, &w);
	int cnt = 0;
	for (int i = 0; i < k; i++)
	{
		for (int j = 0; j < n; j++)
		{
			scanf("%s", s[i] + j*m);
		}
	}
	for (int i = 0; i < k; i++)
	{
		for (int j = i + 1; j < k; j++)
		{
			for (int l = 0; l < n*m; l++)
			{
				dis[i+1][j+1] += s[i][l] != s[j][l];
			}
			dis[i + 1][j + 1] *= w;
			dis[j+1][i+1] = dis[i+1][j+1];
		}
	}
	for (int i = 1; i <= k; i++)
	{
		dis[0][i] = dis[i][0] = n*m;
	}
	for (int j = 1; j <= k; j++)
	{
		low[j] = n*m; ee[j] = edge{ j, 0, n*m };
	}
	int ans = 0;
	for (int i = 0; i < k; i++)
	{
		int index = -1, Min = MAX;
		for (int j = 1; j <= k; j++)
		{
			if (low[j] != -1 && low[j] < Min)
			{
				Min = low[j];
				index = j;
			}
		}
		e[cnt++] = edge{ index, ee[index].y, Min };
		low[index] = -1;
		ans += Min;
		for (int j = 1; j <= k; j++)
		{
			if (low[j] != -1 && low[j]>dis[index][j])
			{
				low[j] = dis[index][j];
				ee[j].y = index;
				ee[j].w = low[j];
			}
		}
	}
	printf("%d\n", ans);
	for (int i = 0; i < cnt; i++)
	{
		printf("%d %d\n", e[i].x, e[i].y);
	}
	
	return 0;
}