昨天刷了一個最簡單的 01 背包，正好趁熱打鐵，多搞幾個背包的題。題目位于 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.
大意就是每組數(shù)據(jù)包含六個值，每個值代表的是擁有該權(quán)重的 marbles 的數(shù)量。求是否可以平均分配。換個角度，這其實就是一個完全背包問題，即是否可以完整裝滿一個 total/2 的背包。
這道題提交了 10 次才 A 掉，4 次 Time Limit Exceeded，5 次 Runtime Error（都是數(shù)組越界造成的）。
下邊的代碼其實依然很亂（估計也沒人看），而且應(yīng)該仍有優(yōu)化空間，不過終究是 A 掉了，先貼出來，稍后在優(yōu)化。
具體優(yōu)化有幾個點：
1、先計算 total，total 為奇數(shù)時必然是不可分的。
2、大于 total/2 的值一律不用計算。因為對結(jié)果沒有影響。
3、先處理奇數(shù)，再處理偶數(shù)。在處理偶數(shù)時，當 total/2 為偶數(shù)時，則只用計算偶數(shù)位，反之則只用計算奇數(shù)位。
4、循環(huán)時可以先計算出當次循環(huán)最大的 index 和最小的 index，過大或者過小的都沒必要處理。
代碼如下：
#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;
// 處理奇數(shù)
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;
}
// 處理偶數(shù)
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");
}
}
}
如下的代碼邏輯比較簡單，但是會超時：
#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");
}
}
}
