POJ 1742 Coins DP 01背包

dp[i][j]表示前i种硬币中取总价值为j时第i种硬币最多剩下多少个，-1表示无法到达该状态。

a.当dp[i-1][j]>=0时，dp[i][j]=ci;

b.当j-ai>=0&&dp[i-1][j-ai]>0时，dp[i][j]=dp[i-1][j-ai]-1;

c.其他，dp[i][j]=-1

Source Code

Problem: 1742        User: BMan
Memory: 1112K        Time: 1547MS
Language: G++        Result: Accepted
Source Code
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\in1.txt","r",stdin);
    freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}


const int maxn=105;
const int maxm=100005;
int c[maxn],a[maxn];
int dp[maxm];

int main()
{
    //freopen("data.in","r",stdin);
    int n,m;
    while(cin>>n>>m)
    {
        if(n&&m);else break;
        for(int i=1;i<=n;i++)
            cin>>a[i];
        for(int i=1;i<=n;i++)
            cin>>c[i];
        memset(dp,-1,sizeof(dp));
        dp[0]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=m;j++)
            {
                if(dp[j]>=0)
                    dp[j]=c[i];
                else if(j>=a[i]&&dp[j-a[i]]>0)
                    dp[j]=dp[j-a[i]]-1;
            }
        }
        int cnt=0;
        for(int i=1;i<=m;i++)
            if(dp[i]>=0)
                cnt++;
        cout<<cnt<<endl;
    }
    return 0;
}