hdu 5445 Food Problem（多重背包）

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题目链接： hdu 5445 Food Problem

解题思路

先对甜点做一次背包，容量表示能量值，对于每个能量值维护最小需要的体积，在对车做一次背包，容量表示花费，对于每个花费值维护最大可以提供的体积。多种背包的优化可以去看一下背包9讲。

代码

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 50000;const int maxm = 205;const int inf = 0x3f3f3f3f;int N, M, P, cost[maxn+5], engy[maxn+5];int T[maxm], U[maxm], V[maxm];int X[maxm], Y[maxm], Z[maxm];void init() {    scanf("%d%d%d", &N, &M, &P);    for (int i = 0; i < N; i++)        scanf("%d%d%d", &T[i], &U[i], &V[i]);    for (int i = 0; i < M; i++)        scanf("%d%d%d", &X[i], &Y[i], &Z[i]);}inline int getbitmost(int x) {    if (x == 0) return 0;    int i = 0;    while (x) {        i++;        x >>= 1;    }    return i-1;}bool solve () {    memset(cost, 0, sizeof(cost));    for (int i = 0; i < M; i++) {        int t = getbitmost(Z[i]);        for (int k = t-1; k >= 0; k--) {            int w = (1<<k) * X[i];            int d = (1<<k) * Y[i];            for (int j = maxn; j >= 0; j--) {                if (j + d > maxn) continue;                cost[j+d] = max(cost[j+d], cost[j]+w);            }        }        int w = (Z[i] - (1<<t) + 1) * X[i];        int d = (Z[i] - (1<<t) + 1) * Y[i];        for (int j = maxn; j >= 0; j--) {            if (j + d > maxn) continue;            cost[j+d] = max(cost[j+d], cost[j]+w);        }    }    memset(engy, inf, sizeof(engy));    engy[0] = 0;    for (int i = 0; i < N; i++) {        int t = getbitmost(V[i]);        for (int k = t-1; k >= 0; k--) {            int w = (1<<k) * U[i];            int d = (1<<k) * T[i];            for (int j = P; j >= 0; j--) {                int v = min(P, j + d);                engy[v] = min(engy[v], engy[j] + w);            }        }        int w = (V[i] - (1<<t) + 1) * U[i];        int d = (V[i] - (1<<t) + 1) * T[i];        for (int j = maxn; j >= 0; j--) {            int v = min(P, j + d);            engy[v] = min(engy[v], engy[j] + w);        }    }    int limit = engy[P];    for (int i = 0; i <= maxn; i++) {        if (limit <= cost[i]) {            printf("%d\n", i);            return true;        }    }    return false;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        if (!solve()) printf("TAT\n");    }    return 0;}

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