POJ 1789 解题报告

来源：互联网发布：知乎上面的神回复编辑：程序博客网时间：2024/05/01 20:14

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。

thestoryofsnow1789Accepted160K438MSC++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 7const 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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