poj 1018 Communication System 枚举 VS 贪心

Communication System

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21631		Accepted: 7689

Description

We have received an order from Pizoor Communications Inc. for a special communication system. The system consists of several devices. For each device, we are free to choose from several manufacturers. Same devices from two manufacturers differ in their maximum bandwidths and prices. 
By overall bandwidth (B) we mean the minimum of the bandwidths of the chosen devices in the communication system and the total price (P) is the sum of the prices of all chosen devices. Our goal is to choose a manufacturer for each device to maximize B/P. 

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. Each test case starts with a line containing a single integer n (1 ≤ n ≤ 100), the number of devices in the communication system, followed by n lines in the following format: the i-th line (1 ≤ i ≤ n) starts with mi (1 ≤ mi ≤ 100), the number of manufacturers for the i-th device, followed by mi pairs of positive integers in the same line, each indicating the bandwidth and the price of the device respectively, corresponding to a manufacturer.

Output

Your program should produce a single line for each test case containing a single number which is the maximum possible B/P for the test case. Round the numbers in the output to 3 digits after decimal point. 

Sample Input

1 3
3 100 25 150 35 80 25
2 120 80 155 40
2 100 100 120 110

Sample Output

0.649

做这题的时候，一开始的想法就是暴力枚举，虽然这样可能会超时，但是觉得可以做强有力的剪枝，于是就试了下，可是没有做论证推断，不知道应该怎样去做剪枝，其实主要是嫌麻烦，下面是我简单的枚举代码，不过超时了。

#include <stdio.h>
#include <stdlib.h>
#define MAX 100

struct Dev{
	int b;
	int p;

}dev[MAX][MAX];

int n;
float result = -1;

void Init()
{
	int i, j;
	for(i=0; i < MAX; i++)
	{
		for(j=0; j < MAX; j++)
		{
			dev[i][j].b= -1;
			dev[i][j].p = -1;
		}
	}
}

void PrintData(int *data, int *sum)
{
	int min=data[0], temp=sum[0], i;
	float tempRes;
	for(i=1; i < n; i++)
	{
		if(data[i] < min) min = data[i];
		temp += sum[i];
	}
	tempRes = min*1.0/ temp;
	if(tempRes > result)
		result = tempRes;
}

void SolveCase(int *data,int *sum, int depth)
{
	int i;
	for(i=0; i<MAX && dev[depth][i].b != -1; i++)
	{
		data[depth] = dev[depth][i].b;
		sum[depth] = dev[depth][i].p;
		if(depth==n-1)
			PrintData(data, sum);
		else
			SolveCase(data,sum, depth+1);
	}
}

int main()
{
//	freopen("input.txt","r",stdin);
	int caseNum, number;
	int *testData, *sum , i, j;

	scanf("%d",&caseNum);
	while(caseNum > 0)
	{
		Init();
		scanf("%d", &n);
		for(i=0; i < n; i++)
		{
			scanf("%d",&number);
			for(j=0; j < number; j++)
				scanf("%d %d", &dev[i][j].b, &dev[i][j].p);
		}
		testData = (int *)malloc(sizeof(int)*n);
		sum = (int *)malloc(sizeof(int)*n);
		SolveCase(testData,sum,0);
		printf("%.3f\n",result);
		result = -1;
		caseNum--;
	}
	free(testData);
	free(sum);
//	fclose(stdin);
	return 0;
}

超时之后，感觉可以用贪心做，然后贪心的话每次使得b 值增大，使得 p 值减少，这样才能使得结果是最大了，思路很简单，以为还是不行，结果AC 了.....

#include<cstdio>
#include<cstring>
int main()
{
//	freopen("input.txt","r",stdin);
	int t,n,m,b[105][105],fac[105],p[105][105],flag[32767],max,min,tp;
	scanf("%d",&t);
	while(t--){
		max=0,min=9999999;
		scanf("%d",&n);
		memset(flag,0,sizeof(flag));
		for(int i=0;i<n;i++){
			scanf("%d",&fac[i]);
			for(int j=0;j<fac[i];j++){
				scanf("%d%d",&b[i][j],&p[i][j]);
				flag[b[i][j]]=1;
				if(max<b[i][j])
					max=b[i][j];
				if(min>b[i][j])
					min=b[i][j];
			}
		}
		double result=0;
		for(int i=min;i<=max;i++){
			if(flag[i]){
				int sum=0;
				for(int j=0;j<n;j++){
					tp=99999999;
					for(int k=0;k<fac[j];k++){
						if(b[j][k]>=i&&p[j][k]<tp){
							tp=p[j][k];
						}
					}
					sum+=tp;
				}
				double temp=(double)i/sum;
				if(result<temp)
					result=temp;
			}
		}
		printf("%.3f\n",result);
	}
//	fclose(stdin);
	return 0;
}