hdu 2660 Accepted Necklace dp / dfs

题目链接

题意：

给出 N 块 石头，从中 至多 选取 k 块（不能重复），重量不得超过 W，求最大价值

思路：

1. 

dp 三重循环 最外层为 当前考虑的石头，里面两层为 取了多少块 和 重量，这两层先后顺序无所谓，总之都是从大到小

然而题意表述不清，以为一定要选取 k 块石头...事实上是至多。

AC代码如下：

#include <cstdio>#include <cstring>#define maxn 22int v[maxn], c[maxn], dp[maxn][1010], n, k;inline max(int a, int b) { return a > b ? a : b; }void work() {    for (int i = 1; i <= n; ++i) scanf("%d%d", &v[i], &c[i]);    int w;    scanf("%d", &w);    memset(dp, 0, sizeof(dp));    for (int i = 1; i <= n; ++i) {        for (int j = k; j >= 1; --j) {            for (int p = w; p >= c[i]; --p) {                dp[j][p] = max(dp[j][p], dp[j - 1][p - c[i]] + v[i]);            }        }    }    printf("%d\n", dp[k][w]);}int main() {    int T;    scanf("%d", &T);    while (scanf("%d%d", &n, &k) != EOF) work();    return 0;}

2.

因为 N <= 22, K <= N, 可以直接 dfs

AC代码如下：

#include <cstdio>#include <cstring>#define maxn 22int v[maxn], c[maxn], dp[maxn][1010], n, k, ans, w;inline max(int a, int b) { return a > b ? a : b; }void dfs(int cnt, int tot, int cost, int val) {    if (tot <= k && cost <= w) {        ans = max(ans, val);        if (tot == k) return;    }    for (int i = cnt + 1; i <= n; ++i) dfs(i, tot + 1, cost + c[i], val + v[i]);}void work() {    for (int i = 1; i <= n; ++i) scanf("%d%d", &v[i], &c[i]);    scanf("%d", &w);    ans = 0;    for (int i = 1; i <= n; ++i) dfs(i, 1, c[i], v[i]);    printf("%d\n", ans);}int main() {    int T;    scanf("%d", &T);    while (scanf("%d%d", &n, &k) != EOF) work();    return 0;}