完全背包和多重背包的混合 F - The Fewest Coins

http://poj.org/problem?id=3260

 

这个题目有点小难，我开始没什么头绪，感觉很乱。

后来看了题解，感觉豁然开朗。

题目大意：就是这个人去买东西，东西的价格是T，这个人拥有的纸币和数量。让你求这个人买东西的纸币量加上老板找给他的纸币量最少是多少。

这个老板用于这个人拥有的纸币种类，数量是无限。

 

思路：

思路就是这个人看成多重背包，老板看成完全背包，f1[i] 表示这个人花了 i 的钱用的最少的纸币。f2[i] 表示老板凑出 i 的钱用的最少的纸币。

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <vector>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 2e5 + 10;
int N, V;
int weight[maxn], num[maxn];
int f1[maxn], f2[maxn], V1;
void zeroonepack(int weight, int val, int f[]) {
    for (int i = V; i >= weight; i--) {
        f[i] = min(f[i], f[i - weight] + val);
    }
}

void completepack(int weight, int val, int f[]) {
    for (int i = weight; i <= V; i++) {
        f[i] = min(f[i], f[i - weight] + val);
    }
}

void multiplepack(int weight, int val, int count, int f[]) {
    if (count*weight >= V) completepack(weight, val, f);
    else {
        int t = 1;
        while (t < count) {
            zeroonepack(weight*t, val*t, f);
            count -= t;
            t *= 2;
        }
        zeroonepack(count*weight, count*val, f);
    }
}

int main() {
    while (scanf("%d%d", &N, &V1) != EOF) {
        int max_val = 0;
        for (int i = 1; i <= N; i++) {
            scanf("%d", &weight[i]);
            max_val = max(max_val, weight[i]);
        }
        for (int i = 1; i <= N; i++) scanf("%d", &num[i]);
        V = max_val * max_val + V1 + 10;
        memset(f1, inf, sizeof(f1));
        memset(f2, inf, sizeof(f2));
        f1[0] = 0, f2[0] = 0;
        for (int i = 1; i <= N; i++) {
            multiplepack(weight[i], 1, num[i], f1);//顾客
        }
        for (int i = 1; i <= N; i++) {
            completepack(weight[i], 1, f2);
        }
        //printf("v=%d v1=%d\n", V, V1);
        int ans = inf;
        for (int i = 0; i <= V - V1; i++) {
            if (f1[V1 + i] != inf && f2[i] != inf) {
                ans = min(f1[V1 + i] + f2[i], ans);
            }
        }
        if (ans != inf) printf("%d\n", ans);
        else printf("-1\n");
    }
    return 0;
}