昨天刷了一个最简单的 01 背包，正好趁热打铁，多搞几个背包的题。题目位于 Dividing，copy 如下：
Description
Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value. Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
Input
Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , n6 , where ni is the number of marbles of value i. So, the example from above would be described by the input-line "1 0 1 2 0 0". The maximum total number of marbles will be 20000.
The last line of the input file will be "0 0 0 0 0 0"; do not process this line.
Output
For each collection, output "Collection #k:", where k is the number of the test case, and then either "Can be divided." or "Can't be divided.".
Output a blank line after each test case.
Sample Input
1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0
Sample Output
Collection #1:
Can't be divided.
Collection #2:
Can be divided.
大意就是每组数据包含六个值，每个值代表的是拥有该权重的 marbles 的数量。求是否可以平均分配。换个角度，这其实就是一个完全背包问题，即是否可以完整装满一个 total/2 的背包。
这道题提交了 10 次才 A 掉，4 次 Time Limit Exceeded，5 次 Runtime Error（都是数组越界造成的）。
下边的代码其实依然很乱（估计也没人看），而且应该仍有优化空间，不过终究是 A 掉了，先贴出来，稍后在优化。
具体优化有几个点：
1、先计算 total，total 为奇数时必然是不可分的。
2、大于 total/2 的值一律不用计算。因为对结果没有影响。
3、先处理奇数，再处理偶数。在处理偶数时，当 total/2 为偶数时，则只用计算偶数位，反之则只用计算奇数位。
4、循环时可以先计算出当次循环最大的 index 和最小的 index，过大或者过小的都没必要处理。
代码如下：
#include <stdio.h>
#include<iostream>
using namespace std;
int MAX_LENGTH = 60000;
int n[7];
bool result[60002];
#define min(a, b) (a < b ? a : b)
int maxSum = 1;
int curTotal = 1;
int dividNum = 1;
// 处理奇数
void processOdd(int value) {
int number = n[value];
int minLoop = (curTotal - dividNum > 1 ? curTotal - dividNum : 1);
int maxLoop = min(maxSum, dividNum);
int tmp = 0;
int k = 0;
for (int t = maxLoop; t >= minLoop; t--) {
if (result[t]) {
for (k = number; k >= 1; k--) {
tmp = t + k * value;
if (tmp <= dividNum) {
result[tmp] = true;
if (maxSum < tmp) {
maxSum = tmp;
}
}
}
}
}
for (int i = 1; i <= number; i++) {
if (value * i <= dividNum) {
result[value * i] = true;
if (maxSum < value * i) {
maxSum = value * i;
}
}
}
curTotal += value * number;
}
// 处理偶数
void processEven(int value, bool isOdd) {
int number = n[value];
int minLoop = (curTotal - dividNum > 1 ? curTotal - dividNum : 1);
int maxLoop = min(maxSum, dividNum);
int tmp = 0;
int k = 0;
if ((maxLoop%2) != isOdd) {
maxLoop -= 1;
}
for (int t = maxLoop; t >= minLoop; t-=2) {
if (result[t]) {
for (k = number; k >= 1; k--) {
tmp = t + k * value;
if (tmp <= dividNum) {
result[tmp] = true;
if (maxSum < tmp) {
maxSum = tmp;
}
}
}
}
}
if (!isOdd) {
for (int i = 1; i <= number; i++) {
if (value * i <= dividNum) {
result[value * i] = true;
if (maxSum < value * i) {
maxSum = value * i;
}
}
}
}
curTotal += value * number;
}
void processResult(int num) {
memset(result, 0, sizeof(result));
maxSum = 1;
curTotal = 1;
dividNum = num;
for (int i = 1; i <=6; i+=2) {
if (n[i] != 0) {
processOdd(i);
}
}
for (int i = 2; i <= 6; i+=2) {
if (n[i] != 0) {
processEven(i, num%2 == 1);
}
}
}
int main(int argc, const char * argv[]) {
int total = 0;
int round = 0;
while (true) {
total = 0;
round++;
for (int i = 1; i <= 6; i++) {
scanf("%d", &n[i]);
total += (n[i] * i);
}
if (total == 0) {
return 0;
}
printf("Collection #%d:\n", round);
if (total % 2 != 0) {
printf("Can't be divided.\n\n");
continue;
}
int dividNum = total / 2;
processResult(dividNum);
if (!result[dividNum]) {
printf("Can't be divided.\n\n");
} else {
printf("Can be divided.\n\n");
}
}
}
如下的代码逻辑比较简单，但是会超时：
#include <stdio.h>
#include<iostream>
using namespace std;
int MAX_LENGTH = 60000;
int n[7];
bool result[60002];
#define min(a, b) (a < b ? a : b)
void processResult(int dividNum) {
int maxSum = 1;
int curTotal = 1;
for (int i = 1; i <= 6; i++) {
if (n[i] != 0) {
for (int k = 1; k <= n[i]; k++) {
int minLoop = (curTotal - dividNum > 1 ? curTotal - dividNum : 1);
for (int t = min(maxSum, dividNum); t >= minLoop; t--) {
if (result[t] && t + i <= dividNum) {
result[t + i] = true;
if (maxSum < t + i) {
maxSum = t + i;
}
}
}
curTotal += i;
if (i*k <= dividNum) {
result[i*k] = true;
if (maxSum < i*k) {
maxSum = i*k;
}
}
}
}
}
}
int main(int argc, const char * argv[]) {
int total = 0;
int round = 0;
while (true) {
total = 0;
round++;
for (int i = 1; i <= 6; i++) {
scanf("%d", &n[i]);
total += (n[i] * i);
}
if (total == 0) {
return 0;
}
printf("Collection #%d:\n", round);
if (total % 2 != 0) {
printf("Can't be divided.\n\n");
continue;
}
memset(result, 0, sizeof(result));
int dividNum = total / 2;
processResult(dividNum);
if (!result[dividNum]) {
printf("Can't be divided.\n\n");
} else {
printf("Can be divided.\n\n");
}
}
}
