HDU - 3535 AreYouBusy(混合背包)

来源：互联网发布：ctf flag.php 编辑：程序博客网时间：2024/05/18 00:56

题目大意：有N类人物，给出每类任务的限制
限制为0时：表示至少要完成1项
限制为1时：表示至多完成1项
限制为2时：可以随意完成

问在给定的限制能，完成这些任务所能得到的最大价值

解题思路：混合背包裸题

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 110;int dp[N][N];int n, m;void solve() {    memset(dp, -1, sizeof(dp));    dp[0][0] = 0;    int mark, num, val, cost;    for (int set = 1; set <= n; set++) {        scanf("%d%d", &num, &mark);        //一个任务都不完成        if (mark != 0)  {            for (int i = 0; i <= m; i++)                dp[set][i] = dp[set - 1][i];        }        while (num--) {            scanf("%d%d", &cost, &val);            if (mark == 0) {                for (int j = m; j >= cost; j--) {                    if (dp[set][j - cost] != -1)                        dp[set][j] = max(dp[set][j], dp[set][j - cost] + val);                    if (dp[set - 1][j - cost] != -1)                        dp[set][j] = max(dp[set][j], dp[set - 1][j - cost] + val);                }            }            else if (mark == 1) {                for (int j = m; j >= cost; j--) {                    if (dp[set - 1][j - cost] != -1)                        dp[set][j] = max(dp[set][j], dp[set - 1][j - cost] + val);                }            }            else {                for (int j = m; j >= cost; j--) {                    if (dp[set][j - cost] != -1)                         dp[set][j] = max(dp[set][j], dp[set][j - cost] + val);                }            }        }    }    int ans = -1;    for (int i = 0; i <= m; i++)        ans = max(ans, dp[n][i]);    printf("%d\n", ans);}int main() {    while (scanf("%d%d", &n, &m) != EOF) solve();    return 0;}

0 0