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;
}