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     0x3f3f3f3ftypedef 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;}