The Fewest Coins POJ

题目传送门

题意：你去一个商店购买一个价值为T的物品，市场上有n种流通的硬币，面值分别为V1,V2..Vn (1<=Vi<=120)。John有Ci个面值为Vi的硬币(0<=Ci<=10000)。店主有无限多的硬币，并总按最优方案找零。
问你最少使用多少个硬币（包括店主找零的）

思路：完全背包加多重背包。

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>#define MAXN 200010
#define MAXE 210
#define INF 1000000000
#define MOD 10001
#define LL long long
#define ULL unsigned long long
#define pi 3.14159using namespace std;int dp[MAXN];
int v[MAXE];
int w[MAXE];
int vis[MAXN];int main() {std::ios::sync_with_stdio(false);int n, m;cin >> n >> m;int max_v = 0;for (int i = 0; i < n; ++i) {cin >> v[i];max_v = max(max_v, v[i]);v[i + n] = -v[i];}for (int i = 0; i < n; ++i) {cin >> w[i];}memset(dp, -1, sizeof(dp));dp[0] = 0;for (int i = 0; i < n; ++i) {memset(vis, 0, sizeof(vis));for (int j = v[i]; j <= m + 30000; ++j) {if (dp[j - v[i]] != -1 && (dp[j - v[i]] + 1 < dp[j] || dp[j] == -1) && vis[j - v[i]] < w[i]) {dp[j] = dp[j - v[i]] + 1;vis[j] = dp[j - v[i]] + 1;}}}for (int i = n; i < 2 * n; ++i) {for (int j = m + 30000 + v[i]; j >= 0; --j) {if (dp[j - v[i]] != -1 && (dp[j - v[i]] + 1 < dp[j] || dp[j] == -1)) {dp[j] = dp[j - v[i]] + 1;}}}cout << dp[m] << endl;return 0;
}

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