POJ 1789 Truck History

来源：互联网发布：淘宝客网站是什么编辑：程序博客网时间：2024/06/10 03:10

题意：求出从这若干序列中选出一个生成其他序列，所需变化花费最小的序列

链接：http://poj.org/problem?id=1789

思路：根据公式可知，要求的d的和就是生成最小树的权值，建图时注意将每一个序列作为一个节点，两个节点之间的权值为需要变化字母的个数

注意点：无

以下为AC代码：

Run IDUserProblemResultMemoryTimeLanguageCode LengthSubmit Time14176748luminous111789Accepted16476K547MSG++1847B2015-05-09 09:28:34

/* ************************************************# @Author  : Luminous11 (573728051@qq.com)*# @Date    : 2015-04-20 18:38:59*# @Link    : http://blog.csdn.net/luminous11                          *********************************************** */#include<cstdio>#include<iostream>#include<string>#include<cstring>#define clr(a, v) memset( a , v , sizeof(a) )using namespace std;const double eps = 1e-10;//const double pi = acos(-1.0);int g[2005][2005];char str[2005][10];int prim ( int n ){    int ans = 0;    int dis[2005];    int vis[2005] = { 0 };    for ( int i = 0; i < n; i ++ ){        dis[i] = g[0][i];    }    for ( int k = 0; k < n; k ++ ){        int _min = 0x3f3f3f3f;        int p = -1;        for ( int i = 0; i < n; i ++ ){            if ( ! vis[i] && _min > dis[i] ){                _min = dis[i];                p = i;            }        }        if ( p == -1 )return -1;        vis[p] = 1;        ans += dis[p];        for ( int i = 0; i < n; i ++ ){            if ( ! vis[i] && dis[i] > g[p][i] ){                dis[i] = g[p][i];            }        }    }    return ans;}int main(){    int n;    while ( cin >> n && n ){        clr ( g, 0 );        for ( int i = 0; i < n; i ++ ){            scanf ( "%s", str[i] );        }        for ( int i = 0; i < n; i ++ ){            for ( int j = i; j < n; j ++ ){                int len = 0;                for ( int k = 0; k < 7; k ++ ){                    if ( str[i][k] != str[j][k] ){                        len ++;                    }                }                g[i][j] = g[j][i] = len;            }        }        int ans = prim( n );        cout << "The highest possible quality is 1/" << ans << "." << endl;    }    return 0;}

0 0