HDOJ-2159 FATE

这道题是二维费用的完全背包问题.两种费用支付的最大值分别为,忍耐度和最多杀怪数.
状态转移方程为: dp[j][h] = max(dp[j][h], dp[j-pa[i]][h-1] + expe[i]);

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>using namespace std;int dp[105][105];int expe[105], pa[105];int main(){   // freopen("in.txt", "r", stdin);    int n, m, k, s;    while(cin >> n >> m >> k >> s)    {        for(int i = 0; i < k; i++)            cin >> expe[i] >> pa[i];        memset(dp, 0, sizeof(dp));        for(int i = 0; i < k; i++)            for(int j = pa[i]; j <= m; j++)            for(int h = 1; h <= s; h++)            dp[j][h] = max(dp[j][h], dp[j-pa[i]][h-1] + expe[i]);        int maxs = -1;        for(int i = 0; i <= m; i++)            for(int j = 0; j <= s; j++)            {                if(dp[i][j] >= n)                    maxs = max(maxs, m - i);            }        cout << maxs << endl;    }    return 0;}