LA_3635/HDU_2333 Assemble (二分)

題意：
讓你去購買不同總類的配件個一個，配件總數爲n，購買的配件總價格不能超過b，且使得所購買的配件中的質量係數中最小的最大
分析：
最小最大的問題，二分枚舉能不能構成不小於k的質量係數,若能構成則ans >= k，否則ans < k.
Code:
#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
#include <vector>
using namespace std;

#define MAXCHAR 20 + 10
#define MAXN    1000 + 10
#define INF     0x3f3f3f3f

typedef struct COMP {
        int q, p;
}Comp;

map<string, int> k;
vector<Comp> components[MAXN];
char k_name[MAXCHAR], name[MAXCHAR];
int k_idx, s[MAXN];

int get_k_idx(char *str)
{
        string tmp;
        tmp.clear();
        for(int i = 0; i < strlen(str); i ++) {
                tmp += str[i];  
        }
        map<string, int>::iterator it;
        it = k.find(tmp);
        if( it != k.end() ) {
                return it->second;
        }
        k.insert(map<string, int>::value_type(tmp, k_idx));
        return k_idx ++;
}


bool valid(int cur_q, int n, int price)
{
        int sum = 0;
        for(int i = 1; i < k_idx; i ++) {
                int p = INF;
                for(int j = 0; j < components[i].size(); j ++) {
                        if( components[i][j].q < cur_q ) {
                                continue;
                        }
                        p = min(p, components[i][j].p);
                }
                if( INF == p ) {
                        return false;
                }
                sum += p;
        }
        if( sum > price ) {
                return false;
        }
        return true;
}


int main(int argc, char **argv)
{
#ifndef ONLINE_JUDGE
        freopen("test.in", "r", stdin);
#endif
        Comp tmp;
        int cas, n, b, max_q;
        scanf("%d", &cas);
        for( ; cas; cas --) {
                scanf("%d %d", &n, &b);
                k.clear();
                for(int i = 1; i <= n; i ++) {
                        components[i].erase(components[i].begin(), components[i].end());
                }
                max_q = 0, k_idx = 1;
                for(int i = 1; i <= n; i ++) {
                        scanf("%s %s %d %d", k_name, name, &tmp.p, &tmp.q);
                        int k = get_k_idx(k_name);
                        components[k].push_back(tmp);
                        max_q = max(max_q, tmp.q);
                }
                
                int L = 0, R = max_q, M, ans = 0;
                while( L < R ) {
                        M = (L+R)/2;
                        if( true == valid(M, n, b) ) {
                                ans = max(ans, M);
                                L = M+1;
                                continue;
                        }
                        R = M;
                }
                if( true == valid(max_q, n, b) ) {
                        ans = max(ans, max_q);
                }
                printf("%d\n", ans);
        }
        return 0;
}