[hdu5445 Food Problem]多重背包

题意：一堆食物，有价值、空间、数量三种属性，一些卡车，有空间，价格，数量三种属性。求最少的钱（不超过50000）买卡车装下价值大于等于给定价值的食物，食物可以拆开来放。

思路：这题的关键是给定的条件：食物可以拆开来放。这个条件使得卡车和食物可以分开考虑，然后通过空间这个属性联系在一起。做两遍多重背包即可。

 

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#pragma comment(linker, "/STACK:10240000")
#include <bits/stdc++.h>
using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;

namespace Debug {
void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<" ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
}
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
/* -------------------------------------------------------------------------------- */

int dp[60000];

void zeroPack_min(int V, int v, int w) {
    for (int i = V; i >= v; i --) {
        umin(dp[i], dp[i - v] + w);
    }
}
void zeroPack_max(int V, int v, int w) {
    for (int i = V; i >= v; i --) {
        umax(dp[i], dp[i - v] + w);
    }
}
void MultiPack(int V, int v, int w, int remain, bool id) {
    if (!remain) return;
    int num = 1;
    while (remain) {
        if (!id) zeroPack_min(V, v * num, w * num);
        else zeroPack_max(V, v * num, w * num);
        remain -= num;
        num <<= 1;
        umin(num, remain);
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T, n, m, p, v, w, c;
    cin >> T;
    while (T --) {
        cin >> n >> m >> p;
        fillchar(dp, 0x3f);
        dp[0] = 0;
        for (int i = 0; i < n; i ++) {
            scanf("%d%d%d", &v, &w, &c);
            MultiPack(p + 100, v, w, c, 0);
        }
        int place = 0x3f3f3f3f;
        for (int i = p; i <= p + 100; i ++) umin(place, dp[i]);
        fillchar(dp, 0);
        for (int i = 0; i < m; i ++) {
            scanf("%d%d%d", &w, &v, &c);
            MultiPack(50000, v, w, c, 1);
        }
        int ans = -1;
        for (int i = 0; i <= 50000; i ++) {
            if (dp[i] >= place) {
                ans = i;
                break;
            }
        }
        if (~ans) cout << ans << endl;
        else puts("TAT");
    }
}
posted @ 2015-09-14 22:58  jklongint  阅读(305)  评论(0)  编辑  收藏  举报