UVA-11517 Exact Change(DP)

题目大意：有n张钞票，面值可能不同。你要买一件东西，可能需要找零钱。问最少付多少钱，并求出最少的钞票张数。

题目分析：定义状态dp(i,w)表示前i张钞票凑成w元需要的最少钞票张数。则状态转移方程为dp(i,w)=min(dp(i-1,w),dp(i-1,w-a(i))+1)。其中a(i)为第i张钞票的面值。

代码如下：

//# define AC

# ifndef AC

# include<iostream>
# include<cstdio>
# include<cstring>
# include<vector>
# include<queue>
# include<list>
# include<cmath>
# include<set>
# include<map>
# include<string>
# include<cstdlib>
# include<algorithm>
using namespace std;

# define mid (l+(r-l)/2)

typedef long long LL;
typedef unsigned long long ULL;

const int N=10000;
const int mod=1e9+7;
const int INF=0x3fffffff;
const LL oo=0x7fffffffffffffff;

int n,w,a[105];
int dp[105][N*2+5];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&w,&n);
        int m=0;
        for(int i=0;i<n;++i){
            scanf("%d",a+i);
            m+=a[i];
        }
        m=min(m,N<<1);
        fill(dp[0],dp[0]+m+1,INF);
        dp[0][0]=0;
        dp[0][a[0]]=1;
        for(int i=1;i<n;++i){
            dp[i][0]=0;
            for(int j=1;j<=m;++j){
                if(j<a[i]) dp[i][j]=dp[i-1][j];
                else{
                    dp[i][j]=INF;
                    if(dp[i-1][j]!=-INF)
                        dp[i][j]=dp[i-1][j];
                    if(dp[i-1][j-a[i]]!=INF)
                        dp[i][j]=min(dp[i][j],dp[i-1][j-a[i]]+1);
                }
            }
        }
        while(dp[n-1][w]==INF||!dp[n-1][w])
            ++w;
        printf("%d %d\n",w,dp[n-1][w]);
    }
    return 0;
}

# endif