HDOJ 2159 FATE

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题意：给出忍耐度和杀怪数量和升级所需经验，并且已知每只怪的经验和所需的忍耐值，求出是否能升级

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2159

思路：有忍耐度和杀怪数量两个限制条件，二维背包，并且每种怪物没有数量限制，完全背包，综合就是二维完全背包。状态转移方程为dp[j][l] = max ( dp[j][l], dp[j-num[i].b][l-1] + num[i].a )

注意点：没说怪物的数量，不能当做01背包

以下为AC代码：

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor123170172014-11-27 00:03:40Accepted215946MS440K1394 BG++luminous11

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>using namespace std;struct node{    int a;    int b;}num[105];int dp[105][105];int main(){    int m, n, s, k;    while ( cin >> n >> m >> k >> s )    {        for ( int i = 0; i < k; i ++ )        {            cin >> num[i].a >> num[i].b;        }        memset ( dp, 0, sizeof ( dp ) );        for ( int i = 0; i < k; i ++ )        {            for ( int j = num[i].b; j <= m; j ++ )            {                for ( int l = 1; l <= s; l ++ )                {                    dp[j][l] = max ( dp[j][l], dp[j-num[i].b][l-1] + num[i].a );                }            }        }        bool flag = 0;        for ( int i = 0; i <= m && ! flag; i ++ )        {            for ( int j = 0; j <= s && ! flag; j ++ )            {                if ( dp[i][j] >= n )                {                    flag = 1;                    cout << m - i << endl;                }            }        }        if ( ! flag )        {            cout << "-1" << endl;        }    }    return 0;}

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