HDU - 1114 Piggy-Bank（完全背包）

题目大意：有一个小猪存钱罐，给出空和满时候的重量，并给出若干种面额的硬币大小和重量，个数不限，问装满小猪时，硬币最小总额是多少，若无法装满，输出impossible

解题思路：完全背包，dp 数组开小了 TLE 好几发……看了半天不知道哪里有问题 还以为会是越界什么的 居然是 TLE

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<string.h>
#include<string>
#include<queue>
#include<map>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
const int INF = 0x3f3f3f3f;
const int NINF = -INF -1;
const int MAXN = 500+5;
using namespace std;
struct coin {
int p, w;
};
coin c[MAXN];
int dp[10005];
int T, E, F, val, n;
int main() {
scanf("%d", &T);
while (T--) {
memset(dp, 0, sizeof(dp));
scanf("%d%d", &E, &F);
val = F - E;
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d%d", &c[i].p, &c[i].w);

for (int i = 0; i <= val; i++)
dp[i] = INF;
dp[0] = 0;
for (int i = 0; i < n; i++)
for (int j = c[i].w; j <= val; j++)
dp[j] = min(dp[j], dp[j-c[i].w]+c[i].p);
//      for (int i = 0; i <= val; i++)
//          printf("%d\n", dp[i]);
if (dp[val] == INF) printf("This is impossible.\n");
else printf("The minimum amount of money in the piggy-bank is %d.\n", dp[val]);

}
return 0;
}