[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?:-;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[]; 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 = ; while (remain) { if (!id) zeroPack_min(V, v * num, w * num); else zeroPack_max(V, v * num, w * num); remain -= num; num <<= ; 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[] = ; for (int i = ; i < n; i ++) { scanf("%d%d%d", &v, &w, &c); MultiPack(p + , v, w, c, ); } int place = 0x3f3f3f3f; for (int i = p; i <= p + ; i ++) umin(place, dp[i]); fillchar(dp, ); for (int i = ; i < m; i ++) { scanf("%d%d%d", &w, &v, &c); MultiPack(, v, w, c, ); } int ans = -; for (int i = ; i <= ; i ++) { if (dp[i] >= place) { ans = i; break; } } if (~ans) cout << ans << endl; else puts("TAT"); } }