FATE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11908 Accepted Submission(s): 5645
題目鏈接:HDU 2159
比較簡單基礎二維揹包,跟一維一個道理,最後回去找一下最小忍耐消耗的即可
代碼:
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=110;
int c[N],d[N],w[N],dp[N][N];
int n,m,k,s;
void init()
{
MM(c,0);
MM(d,0);
MM(w,0);
MM(dp,0);
}
int main(void)
{
int i,j,z;
while (~scanf("%d%d%d%d",&n,&m,&k,&s))
{
init();
for (i=0; i<k; ++i)
{
scanf("%d%d",w+i,c+i);
}
int minm=INF;
for (i=0; i<k; ++i)
{
for (j=c[i]; j<=m; ++j)
{
for (z=1; z<=s; ++z)
{
dp[j][z]=max(dp[j][z],dp[j-c[i]][z-1]+w[i]);
}
}
}
if(dp[m][s]<n)
puts("-1");
else
{
int minm=INF;
for (i=0; i<=m; ++i)//忍耐度
for(j=0; j<=s; ++j)//殺怪數量
if(dp[i][j]>=n&&i<minm)
minm=i;
printf("%d\n",m-minm);
}
}
return 0;
}