【UVA】1395-Slim Span

来源：互联网发布：普通话水平测试软件编辑：程序博客网时间：2024/06/05 07:26

数学实在练不下去了，只能来水几个图论了，真想像D神一样来句：这道题很简单，直接AC就可以了。

大体思路：按照边的权值排序，枚举区间，利用并查集判断是否构成通路。

140426631395Slim SpanAcceptedC++0.2652014-08-15 02:11:53

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<stack>#include<queue>#include<map>#include<set>#include<list>#include<cmath>#include<string>#include<sstream>#include<ctime>using namespace std;#define _PI acos(-1.0)#define INF (1 << 10)#define esp 1e-9typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*======================================================================================*/#define MAXD 5000 + 10#define MAX_SIZE 100 + 10int fa[MAX_SIZE];struct Line{    int x;    int y;    int len;    friend bool operator <(Line p,Line q){         if(p.len < q.len)            return true;         else            return false;    }}node[MAXD];int n,m;int find_father(int u){    return fa[u] == u ? u : fa[u] = find_father(fa[u]);}void init(){    for(int i = 1 ; i <= n ; i++)        fa[i] = i;}bool Judge(){    int root = find_father(1);  /*找到1的父亲*/    for(int i = 2 ; i <= n ; i++){        int _node = find_father(i);        if(_node != root)            return false;    }    return true;}int main(){    while(scanf("%d%d",&n,&m)){        if(!n && !m)            break;        for(int i = 0 ; i < m ; i++)            scanf("%d%d%d",&node[i].x,&node[i].y,&node[i].len);        sort(node,node + m);        int ans = -1;        for(int i = 0 ; i < m ; i++){            init();            for(int j = i ; j < m ; j++){                int x = node[j].x;                int y = node[j].y;                int _x = find_father(x); /*找到x的父亲*/                int _y = find_father(y); /*找到y的父亲*/                if(_x != _y)                fa[_x] = _y;                if(Judge()){                    if(ans >= 0)                    ans = min(ans,node[j].len - node[i].len);                    else                    ans = node[j].len - node[i].len;                    break;                }            }        }        printf("%d\n",ans);    }    return 0;}

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