POJ3497 UVA12124 Assemble(二分 + 贪心)

Assemble

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3361 Accepted: 1070

Description

Recently your team noticed that the computer you use to practice for programming contests is not good enough anymore. Therefore, you decide to buy a new computer.

To make the ideal computer for your needs, you decide to buy separate components and assemble the computer yourself. You need to buy exactly one of each type of component.

The problem is which components to buy. As you all know, the quality of a computer is equal to the quality of its weakest component. Therefore, you want to maximize the quality of the component with the lowest quality, while not exceeding your budget.

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

• One line with two integers: 1 ≤ n ≤ 1 000, the number of available components and 1 ≤ b ≤ 1 000 000 000, your budget.

• n lines in the following format: “type name price quality”, where type is a string with the type of the component, name is a string with the unique name of the component, price is an integer (0 ≤ price ≤ 1 000 000) which represents the price of the component and quality is an integer (0 ≤ quality ≤ 1 000 000 000) which represents the quality of the component (higher is better). The strings contain only letters, digits and underscores and have a maximal length of 20 characters.

Output

Per testcase:

• One line with one integer: the maximal possible quality.

Sample Input

118 800processor 3500_MHz 66 5processor 4200_MHz 103 7processor 5000_MHz 156 9processor 6000_MHz 219 12memory 1_GB 35 3memory 2_GB 88 6memory 4_GB 170 12mainbord all_onboard 52 10harddisk 250_GB 54 10harddisk 500_FB 99 12casing midi 36 10monitor 17_inch 157 5monitor 19_inch 175 7monitor 20_inch 210 9monitor 22_inch 293 12mouse cordless_optical 18 12mouse microsoft 30 9keyboard office 4 10

Sample Output

9

Source

Northwestern Europe 2007

解题思路:

最小值最大问题,可以考虑用二分答案来做,二分物品的品质，删除品质小于当前二分的值,如果可以组装一台电脑那么答案ans >= x,否则ans < x，本题对编程和代码规范要求还是比较高的，值得一做。期间因为vector没清空，RE了一次,还有不能用cin,必须用scanf,效率真是差得非常多。

AC代码:

#include<iostream>#include<string>#include<map>#include<vector>#include<algorithm>#include<cstdio>using namespace std;const int maxn = 1005;const int inf = 0x3f3f3f3f;const int word = 25;int cnt; //记录不同总类的ID int m,n;map<string,int> ID;struct com{int cost;int quality;};int Max;vector<com> comp[maxn]; void init(){cnt = 0;int i;Max = -inf;for(i=0;i<m;i++){comp[i].clear();}ID.clear();}int getTypeId(string name){if(!ID.count(name)) ID[name] = cnt++;return ID[name];}bool ok(int quality) //求最小的品质 {int i,j;int temp;int cost = 0;com g,f;int cheap;for(i=0;i<cnt;i++) //每个物品 {cheap = inf;int Size = (int)comp[i].size(); //遍历每个物品的件数 //cout<<"Size:"<<Size<<endl;for(j=0;j<Size;j++){g = comp[i][j];temp = g.quality;if(temp >= quality){cheap = min(g.cost,cheap); }}if(cheap == inf) {//cout<<"----------"<<endl;return false;}cost += cheap;//cout<<cost<<endl;if(cost > n){//cout<<"--------"<<endl;return false;}}return true;}int main(){int T;int i;//freopen("1.txt","r",stdin);scanf("%d",&T);int value,qu;char str1[word],str2[word];while(T--){init();scanf("%d%d",&m,&n);for(i=0;i<m;i++){scanf("%s%s%d%d",&str1,&str2,&value,&qu);Max = max(Max,qu);int num = getTypeId(str1);com c;c.cost = value;c.quality = qu;comp[num].push_back(c); //第几个物品对应的价格和品质 }int left,right;int mid;left = 0,right = Max;//cout<<Max<<endl;while(left < right){mid = left + (right - left + 1) / 2;if(ok(mid)){left = mid;//cout<<"left:"<<left<<endl;}else{right = mid - 1;//cout<<"right:"<<right<<endl;}}printf("%d\n",left);}return 0;}