[2016-03-19][UVALive][3971][Assemble]

• 时间:2016-03-19 13:55:17 星期六

• 题目编号:[2016-03-19][UVALive][3971][Assemble]

• 题目大意:给定若干个电脑零件的价格和质量,求在总价不超过b的情况下,品质最差的配件的质量尽可能大

• 分析:二分

#include <vector>
#include <map>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
#define CLR(x,y) memset((x),(y),sizeof((x)))
#define CLR2(x,y,mtype,mcnt) memset((x),(y),sizeof((mtype))*(mcnt))
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);++(x))
#define FORD(x,y,z) for(int (x)=(y);(x)>=(z);--(x))
const int maxn = 1000 + 100;
struct Com{
        int price,quality;
        Com(int p = 0,int q = 0):price(p),quality(q){}       
};
vector<Com> comp[maxn];
map<string,int> id;
int t,n,b,p,q,maxq,cnt;
int ID(string str){
        if(id.count(str))       return id[str];
        else id[str] = cnt;
        return cnt++;
}
int ok(int q){
        int sum = 0;
        FOR(i,0,cnt){
                int chepest = b + 1;  
                int m = comp[i].size();
                FOR(j,0,m){
                        if(comp[i][j].quality >= q && comp[i][j].price < chepest)
                                chepest = comp[i][j].price;
                }
                sum += chepest;
                if(sum > b)     return 0;
        }
        return 1;
}
int main(){     
        char type[30];
        scanf("%d",&t);
        while(t--){
                scanf("%d%d",&n,&b);
                cnt = 0;
                FOR(i,0,n)      comp[i].clear();
                id.clear();
                maxq = 0;
                FOR(i,0,n){
                        scanf("%s %*s %d %d",type,&p,&q);
                        if(q > maxq)    maxq = q;
                        comp[ID(type)].push_back( Com(p,q) );
                }               
                int l = 0,r = maxq,mid;
                while(l < r){
                        mid = l + (r - l + 1)/2;
                        if(ok(mid))     l = mid;
                        else            r = mid - 1;
                }
                printf("%d\n",l);
        }
        return 0;
}

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