POJ 3260 The Fewest Coin

WA 了三次

1. solve_dp 中, 第一轮 把 cpmoney 写成了 money

2. 第二轮 dp 中, 把 +1 写成 +number[i]

3. T 的最大值设置出了问题

总结

1. 需要在纸上想好再动笔, 写这个代码的时候, solve_dp 的参数修改了若干次, 导致 WA 的次数过多

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <deque>
#include <cstring>
#define MIN(x,y) (x)<(y)?(x):(y)
using namespace std;

int money[200];
int number[200];

int dp[50000];

int cpmoney[2000];
int cpnumber[2000];

int transTo01Pack(int n) {
	int newn = 0;

	for(int i = 0; i < n; i ++)  {
		int num = number[i];
		bool stop_flag = false;
		for(int j = 0; !stop_flag; j ++) {
			if(num - (1<<j) >= 0)  {
				cpmoney[newn] = money[i]*(1<<j);
				cpnumber[newn] = (1<<j);
				newn ++;
				num -= (1<<j);
			}  else {
				stop_flag = true;
			}
		}

		if(num > 0)  {
			cpmoney[newn] = money[i] * num;
			cpnumber[newn] = num;
			newn ++;
		}
	}

	return newn;
}


int solve_dp(int cpn, int n, int cpt, int t)  {
	memset(dp, 0x3f, sizeof(dp));
	// first pass
	// 01 pack
	dp[0] = 0;

	for(int i = 0; i < cpn; i ++)  {
		for(int v = cpt; v >= cpmoney[i]; v --)  {
			dp[v] = min(dp[v], dp[v-cpmoney[i]] + cpnumber[i]);
		}
	}

	// second pass
	// complete pack
	for(int i = 0; i < n; i ++)  {
		for(int v = cpt-money[i]; v >= 0; v --)  {
			dp[v] = min(dp[v], dp[v+money[i]] + 1);
		}
	}

	if(dp[t] == 0x3f3F3F3F)
		return -1;
	return dp[t];
}

int main() {
	freopen("C:\Users\vincent\Dropbox\workplacce\joj\test.txt", "r", stdin);
	
	int N, T;
	while(scanf("%d%d", &N, &T) != EOF)  {
		for(int i = 0; i < N; i ++)  {
			scanf("%d", money+i);
		}
		for(int i = 0; i < N; i ++)  {
			scanf("%d", number+i);
		}
		int cpn = transTo01Pack(N);
		int res = solve_dp(cpn, N, 50000, T);
		printf("%d
", res);
	}

	return 0;
}