UVaLive 3971 Assemble (水题二分+贪心)

题意：你有b元钱，有n个配件，每个配件有各类，品质因子，价格，要每种买一个，让最差的品质因子尽量大。

析：很简单的一个二分题，二分品质因子即可，每次计算要花的钱的多少，每次尽量买便宜且大的品质因子。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

struct Node{
    int p;
    int val;
    bool operator < (const Node &pp) const{
        return p < pp.p;
    }
    Node(int pp, int v) : p(pp), val(v) { }
};
vector<Node> v[maxn];
map<string, int> mp;
int cnt;

int getId(const string &s){
    if(mp.count(s))  return mp[s];
    return mp[s] = cnt++;
}

bool judge(int mid){
    int ans = 0;
    for(int i = 0; i < cnt; ++i){
        bool ok = false;
        for(int j = 0; j < v[i].size(); ++j)
            if(v[i][j].val >= mid){ ans += v[i][j].p;  ok = true; break; }
        if(!ok || ans > m)  return false;
    }
    return true;
}

int solve(){
    int l = 1, r = (int)1e9;
    for(int i = 0; i < cnt; ++i)  sort(v[i].begin(), v[i].end());
    while(l < r){
        int mid = (l + r) >> 1;
        if(judge(mid)) l = mid + 1;
        else r = mid;
    }
    return judge(l) ? l : l-1;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d %d", &n, &m);
        mp.clear();
        cnt = 0;
        for(int i = 0; i < n; ++i)  v[i].clear();
        char s[25], t[25];
        int p, x;
        for(int i = 0; i < n; ++i){
            scanf("%s %s %d %d", s, t, &p, &x);
            v[getId(s)].push_back(Node(p, x));
        }
        printf("%d\n", solve());
    }
    return 0;
}