POJ 1789 解题报告

本文主要是介绍POJ 1789 解题报告，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

这道题是求最小生成树。很久之前是用kruskal算法求的（之前已经用过这个模板很多次），但是超时了，这里是稠密图，对所有边排序是非常耗时的操作。这里改用没有优化的prim算法（用的是数组而不是heap，这意味着每次选最近的节点都需要过一遍数组，O(N)）。但是还是很轻松地通过了。

thestoryofsnow

1789

Accepted

160K

438MS

C++

1591B

/* 
ID: thestor1 
LANG: C++ 
TASK: poj1789 
*/
#include <iostream>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <limits>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <cassert>using namespace std;const int MAXN = 2000;
bool inset[MAXN];
int dis[MAXN];// each line is a 7-character string
// therefore, the maximum difference is 7
const int MAXDIFF = 7;
char lines[MAXN][MAXDIFF + 1];int diff(int u, int v)
{int d = 0;for (int i = 0; lines[u][i] != '\0' && lines[v][i] != '\0'; ++i){if (lines[u][i] != lines[v][i]){d++;}}return d;
}int minu(const int N)
{int mindis = MAXDIFF, u = 0;for (int i = 0; i < N; ++i){if (!inset[i] && dis[i] < mindis){mindis = dis[i];u = i;}}return u;
}int prim(const int N)
{for (int i = 0; i < N; ++i){inset[i] = false;}for (int i = 0; i < N; ++i){dis[i] = MAXDIFF;}int source = 0;dis[source] = 0;int mst = 0;for (int i = 0; i < N; ++i){int u = minu(N);	inset[u] = true;mst += dis[u];for (int i = 0; i < N; ++i){if (!inset[i]){int d = diff(u, i);if (d < dis[i]){dis[i] = d;}}}}return mst;
}int main()
{int N;while (scanf("%d", &N) && N){for (int i = 0; i < N; ++i){scanf("%s", lines[i]);}// for (int i = 0; i < N; ++i)// {// 	printf("%s\n", lines[i]);// }printf("The highest possible quality is 1/%d.\n", prim(N));}return 0;
}

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