uvalive3971

#pragma warning(disable:4786)#include <stdio.h>#include <iostream>#include <string>#include <vector>#include <map>#define zzzusing namespace std;int min(int a, int b){return a<b?a:b;}int max(int a, int b){return a>b?a:b;}const int MAXN = 1000 + 5;int cnt;map<string, int>id;int ID(string s){if(!id.count(s)) id[s]=cnt++;return id[s];}struct ZZ{int p, q;};vector<ZZ>zz[MAXN];int b, n;bool erfen(int q){int sum = 0;for(int i=0; i<cnt; i++){int bargin = b + 1;int m = zz[i].size();for(int j=0; j<m; j++){if(zz[i][j].q>=q) bargin = min(bargin, zz[i][j].p);}if(bargin == b+1) return false;sum += bargin;if(sum>b) return false;}return true;}int main(){#ifndef zzzfreopen("in.txt", "r", stdin);#endifint cas;scanf("%d", &cas);while(cas--){scanf("%d%d", &n, &b);cnt = 0;int i;for(i=0; i<n; i++) zz[i].clear();id.clear();int maxq = 0;for(i=0; i<n; i++){char type[30], name[30];int p, q;scanf("%s%s%d%d", type, name, &p, &q);maxq = max(maxq, q);ZZ tmp;tmp.p = p;tmp.q = q;zz[ID(type)].push_back(tmp);}int l = 0;int r = maxq;while(l<r){int m = (l+r+1)/2;if(erfen(m)) l = m;else r = m - 1;}printf("%d\n", l);}return 0;}

题意：你有b元钱，想要组装一台电脑，给出n个配件各自的种类、品质银子和价格，要求每种类型的配件各买一个，总价格不超过b，且“品质最差配件”的品质因子应该尽量大

做法：如果每种配件都买最差的，当然有解，所以二分所有配件的品质就可以

代码：