HDU 5445 Food Problem 解题报告（背包）

HOT~ 杭电2015级新生如何加入ACM集训队？

Food Problem

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 120    Accepted Submission(s): 33

Problem Description

Few days before a game of orienteering, Bell came to a mathematician to solve a big problem. Bell is preparing the dessert for the game. There are several different types of desserts such as small cookies, little grasshoppers and tiny mantises. Every type of dessert may provide different amounts of energy, and they all take up different size of space.

Other than obtaining the desserts, Bell also needs to consider moving them to the game arena. Different trucks may carry different amounts of desserts in size and of course they have different costs. However, you may split a single dessert into several parts and put them on different trucks, then assemble the parts at the game arena. Note that a dessert does not provide any energy if some part of it is missing.

Bell wants to know how much would it cost at least to provide desserts of a total energy ofp (most of the desserts are not bought with money, so we assume obtaining the desserts costs no money, only the cost of transportation should be considered). Unfortunately the mathematician is having trouble with her stomach, so this problem is left to you.

 

Input

The first line of input contains a integer T(T≤10) representing the number of test cases.

For each test case there are three integers n,m,p on the first line (1≤n≤200,1≤m≤200,0≤p≤50000), representing the number of different desserts, the number of different trucks and the least energy required respectively.

The i−th of the n following lines contains three integers ti,ui,vi(1≤ti≤100,1≤ui≤100,1≤vi≤100) indicating that the i−th dessert can provide ti energy, takes up space of size ui and that Bell can prepare at most vi of them.

On each of the next m lines, there are also three integers xj,yj,zj(1≤xj≤100,1≤yj≤100,1≤zj≤100) indicating that the j−th truck can carry at most size of xj , hiring each one costs yj and that Bell can hire at most zj of them.

 

Output

For every test case output the minimum cost to provide the dessert of enough energy in the game arena if it is possible and its cost is no more than50000. Otherwise, output TAT on the line instead.

 

Sample Input

41 1 714 2 11 2 21 1 1010 10 15 7 25 3 341 4 19 4 25 3 31 3 35 3 23 4 56 7 55 3 81 1 11 2 11 1 1

 

Sample Output

41412TAT

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

    解题报告：背包。首先计算出满足能量值p的最小空间大小，再由最小空间大小背包出最少金钱数。

    简单计算数据，最小空间大小极端情况下可能会是200 * 100 * 100 = 200W，200W背包肯定会超时的。

    注意Output里的提示，钱最多给50000，按照50000的大小背包出钱对应的最大空间，然后求出符合条件的最少的钱即可。

    PS：比赛的时候我和队友根本没看到50000的条件……但是我们AC了……这是一个悲伤的故事

#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <set>#include <map>#include <queue>#include <vector>#include <functional>#include <cassert>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;#define ff(i, n) for(int i=0,END=(n);i<END;i++)#define fff(i, n, m) for(int i=(n),END=(m);i<=END;i++)#define dff(i, n, m) for(int i=(n),END=(m);i>=END;i--)#define travel(e, u) for(int e=first[u], v=vv[first[u]]; ~e; e=nxt[e])#define mid ((l+r)/2)#define bit(n) (1ll<<(n))#define clr(a, b) memset(a, b, sizeof(a))#define debug(x) cout << #x << " = " << x << endl;#define ls (rt << 1)#define rs (ls | 1)#define lson l, m, ls#define rson m + 1, r, rsvoid work();int main() {    work();    return 0;}/**************************Beautiful GEGE**********************************/const int maxn = 222;const int maxv = 5e4 + 5 + maxn;const int inf = 0x3f3f3f3f;int n, m, p;int dp[maxv];void zero_one_pack(int v, int c, int V) {    dff(i, V, v) {        dp[i] = min(dp[i], dp[i - v] + c);    }}void zero_one_pack2(int v, int c, int V) {    dff(i, V, v) {        dp[i] = max(dp[i], dp[i - v] + c);    }}void multi_pack(int v, int c, int num, int V, bool flag = true) {    int k = 1;    while (num) {        if (flag)            zero_one_pack(v * k, c * k, V);        else            zero_one_pack2(v * k, c * k, V);        num -= k;        k += k;        if (k > num) k = num;    }}void input() {    scanf("%d%d%d", &n, &m, &p);    clr(dp, inf); dp[0] = 0;    ff (i, n) {        int v, c, num;        scanf("%d%d%d", &v, &c, &num);        multi_pack(v, c, num, p + 100);    }    int V = *min_element(dp + p, dp + p + 100);    clr(dp, 0);    ff (i, m) {        int v, c, num;        scanf("%d%d%d", &v, &c, &num);        multi_pack(c, v, num, 50000, false);    }    int ans = inf;    dff(i, 50000, 0) if (dp[i] >= V) {        ans = min(ans, i);    }    if (ans == inf) {        puts("TAT");    } else {        printf("%d\n", ans);    }}void work() {    int T; scanf("%d", &T);    fff(cas, 1, T) {        input();    }}