UVA 10099 - The Tourist Guide

来源：互联网发布：安卓4.0淘宝编辑：程序博客网时间：2024/06/15 19:17

题意：有一个导游，要带领团队从当前点，走到目标点，但是每条路一次性通过的人有数量限制，

所以需要分批次进行。问最少需要分多少次进行。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#define ll 0x7fffffffusing namespace std;typedef struct d_node{    int point1;    int point2;    int weight;}enode;enode edge[10001];//边，输入存储//union_setint sets[101];//点的索引int Rank[101];//那个点作为根节点void set_inital( int a, int b ){    for ( int i = a ; i <= b ; ++ i ) {        Rank[i] = 0;        sets[i] = i;    }}int  set_find( int a ){    if ( a != sets[a] )        sets[a] = set_find( sets[a] );    return sets[a];}void Set_union( int a, int b ){    if ( Rank[a] < Rank[b] )        sets[a] = b;    else {        if ( Rank[a] == Rank[b] )            Rank[a] ++;        sets[b] = a;    }}//end_union_setint cmp_e( enode a, enode b ){    return a.weight > b.weight;}int Vector[101];//通过此点的最小权值int kruskal( int n, int r, int s, int t ){    set_inital( 1, n );    for ( int i = 1 ; i <= n ; ++ i )        Vector[i] = ll;        //cout<<vector[1]<<endl;    sort( edge, edge+r, cmp_e );    for ( int i = 0 ; i < r ; ++ i ) {        int A = set_find( edge[i].point1 );        int B = set_find( edge[i].point2 );        if ( A != B ) {            Set_union( A, B );            int mid = min( Vector[A], Vector[B] );            Vector[set_find(A)] = min( mid, edge[i].weight );            if ( set_find( s ) == set_find( t ) )                return Vector[set_find(s)];        }    }    return -1;}int main(){    int N,R,a,b,c,s,t,p,C = 1;    while ( scanf("%d%d",&N,&R) && N+R ) {        for ( int i = 0 ; i < R ; ++ i ) {            scanf("%d%d%d",&a,&b,&c);            edge[i].point1 = a;            edge[i].point2 = b;            edge[i].weight = c;        }        scanf("%d%d%d",&s,&t,&p);        int W = kruskal( N, R, s, t );        int A = p/(W-1) + (p%(W-1) != 0);        printf("Scenario #%d\nMinimum Number of Trips = %d\n\n",C ++,A);    }    return 0;}

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