HDU 1114

一道动归的水题

 

一开始AC的代码：

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define MIN(A,B) (A<B?A:B)
#define LEN 10001
#define MMAX 0x3f3f3f3f

int dp[LEN], mark[LEN];

int main() {
	int t, e, f;
	int n;
	int p, w;
	scanf("%d", &t);
	while(t--) {
		int i, j, tail;
		scanf("%d%d%d", &e, &f, &n);
		for(i = 0; i < LEN; i++) {
			dp[i] = MMAX;
		}
		memset(mark, 0, sizeof(mark));
		mark[0] = 1;
		dp[0] = 0;
		tail = f-e;
		for(i = 0; i < n; i++) {
			scanf("%d%d", &p, &w);
			for(j = w; j <= tail; j++) {
				if(1 == mark[j-w]) {
					dp[j] = MIN(dp[j], dp[j-w]+p);
					mark[j] = 1;
				}
			}
		}
		if(MMAX == dp[tail]) {
			printf("This is impossible.\n");
		}
		else {
			printf("The minimum amount of money in the piggy-bank is %d.\n", dp[tail]);
		}
	}
	return 0;
}

 

后来发现其实不加  if(1 == mark[j-w])  这个判断也能过，因为当他没有被计算过的时候，dp的值就是MMAX，也就会忽略 dp[j] 的大小。

改进后：

 

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define MIN(A,B) (A<B?A:B)
#define LEN 10001
#define MMAX 0x3f3f3f3f

int dp[LEN], mark[LEN];

int main() {
	int t, e, f;
	int n;
	int p, w;
	scanf("%d", &t);
	while(t--) {
		int i, j, tail;
		scanf("%d%d%d", &e, &f, &n);
		for(i = 0; i < LEN; i++) {
			dp[i] = MMAX;
		}
		memset(mark, 0, sizeof(mark));
		mark[0] = 1;
		dp[0] = 0;
		tail = f-e;
		for(i = 0; i < n; i++) {
			scanf("%d%d", &p, &w);
			for(j = w; j <= tail; j++) {
//				if(1 == mark[j-w]) {
					dp[j] = MIN(dp[j], dp[j-w]+p);
//					mark[j] = 1;
//				}
			}
		}
		if(MMAX == dp[tail]) {
			printf("This is impossible.\n");
		}
		else {
			printf("The minimum amount of money in the piggy-bank is %d.\n", dp[tail]);
		}
	}
	return 0;
}