HDU 1114 Piggy-Bank 完全背包

题目大意：

就是现在给出每个硬币的面值和重量，现在一个罐子里面装满了总重量为W的硬币，问最少的价值是多少

大致思路：

首先每个物品的体积就是重量，价值就是面值，要求完全装满，初始化f [ 1 ~ V ] 都为负无穷， f [ 0 ] 为0， 状态转移的时候取较小的值，每个物品可以放任意个故是一个完全背包

代码如下：

Result  :  Accepted     Memory  :  296 KB     Time  :  78 ms

/*
* Author: Gatevin
* Created Time: 2014/8/15 19:09:56
* File Name: haha.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

int f[10010];
int w[510];
int c[510];
int n,E,F,V;
const int inf = 0x7fffffff;
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &E, &F);
V = F - E;
if(V < 0)
{
printf("This is impossible.\n");
continue;
}
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%d %d", &w[i], &c[i]);
}
f[0] = 0;
for(int i = 1; i <= V; i++) f[i] = -inf;
for(int i = 1; i <= n; i++)
{
for(int j = c[i]; j <= V; j++)
{
if(f[j] != -inf && f[j - c[i]] != -inf )
{
f[j] = min(f[j], f[j - c[i]] + w[i]);
continue;
}
if(f[j] == -inf && f[j - c[i]] != -inf)
{
f[j] = f[j - c[i]] + w[i];
continue;
}
}
}
if(f[V] == -inf)
{
printf("This is impossible.\n");
}
else
{
printf("The minimum amount of money in the piggy-bank is %d.\n", f[V]);
}
}
return 0;
}