标签：math.h break include div queue const pass ios names
题意：从A市到B市有n个站点,限制火车上搭乘的乘客数目,每个站点都从有一些乘车的订单,订单信息 从x到y,乘客m人,求解最大的收入是多少
最多7个站,22个订单
选取订单的时候没有顺序问题,所以不是全排列,是组合问题,一个订单俩种情况,拿OR不拿,所以是2^22次方种组合,
pass数组用于记录拿了这些订单后火车在每个站点时的人数
#include<stdio.h>
#include<iostream>
#include<sstream>
#include<queue>
#include<map>
#include<memory.h>
#include <math.h>
#include<time.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
const int N = 110;
class Node
{
public:
int s;
int e;
int num;
Node(int s, int e, int num)
{
this->s = s;
this->e = e;
this->num = num;
}
Node()
{
this->s = 0;
this->e = 0;
this->num = 0;
}
};
int limit;
int station;
int orders;
int pass[9];
Node node[25];
int maxSum = -1;
void read(int orders, Node nodes[])
{
for(int i = 0; i < orders; i++)
{
Node node;
cin >> node.s >> node.e >> node.num;
nodes[i] = node;
}
}
void dfs(int cn, int sum)
{
if(cn == orders)
{
maxSum = sum > maxSum ? sum : maxSum;
return;
}
int ok = 1;
int sn[10];
memset(sn,0,sizeof(sn));
for(int i = node[cn].s; i < node[cn].e; i++)
if(pass[i] + node[cn].num <= limit)
{
pass[i] += node[cn].num;
sn[i] = 1;
}
else
{
ok = 0;
break;
}
if(ok)
dfs(cn + 1, (node[cn].e - node[cn].s) * node[cn].num + sum);
for(int i = node[cn].s; i < node[cn].e; i++)
if(sn[i])
pass[i] -= node[cn].num;
dfs(cn + 1, sum);
}
int main()
{
freopen("d:\\1.txt", "r", stdin);
while ((cin >> limit >> station >> orders) && limit)
{
memset(pass, 0, sizeof(pass));
maxSum = -1;
read(orders, node);
dfs(0,0);
cout<<maxSum<<endl;
}
return 0;
}
标签：math.h break include div queue const pass ios names
原文地址：http://www.cnblogs.com/shuiyonglewodezzzzz/p/7751251.html
