hdu-1114 Piggy-Bank---完全背包

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1114

题目大意：

给出小猪钱罐的重量和装满钱后的重量，然后是几组数据，每组数据包括每种钱币的价值与重量

要求出重量最少能装满钱罐时的最大价值

思路：

完全背包裸题，dp[j] = min(dp[j], dp[j-w[i]]+v[i])

注意dp数组开的范围，一开始开小了，一直WA

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
#include<map>
#include<cmath>
#include<sstream>
using namespace std;
typedef pair<int, int> Pair;
typedef long long ll;
const int INF = 0x3f3f3f3f;
int T, n, m, minans;
const int maxn = 1e3 + 10;
int dir[4][2] = {1,0,0,1,-1,0,0,-1};
int w[maxn], v[maxn];
int dp[100000];
int main()
{
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        m -= n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)scanf("%d%d", &v[i], &w[i]);
        memset(dp, INF, sizeof(dp));
        dp[0] = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = w[i]; j <= m; j++)
                dp[j] = min(dp[j], dp[j - w[i]] + v[i]);
        }
        if(dp[m] >= 1000000)
            printf("This is impossible.
");
        else printf("The minimum amount of money in the piggy-bank is %d.
", dp[m]);
    }
    return 0;
}