ZOJ 3331 Process the Tasks（双塔DP）

Process the Tasks

Time Limit: 1 Second Memory Limit: 32768 KB

There are two machines A and B. There are n tasks, namely task 1, task 2, ..., task n. You must assign each task to one machine to process it. There are some facts you
must know and comply with:

You must assign each task to only one machine to process.
At any time, a machine can process at most one task.
Task i (0 < i < n) can be processed if and only if each task j (0 < j < i) has been processed or processing.
If a task is processing on one machine, it cannot be interrupted.

You want to do finish all the tasks as soon as possible.

Input

There are multiple test cases. The first line of the input is an integer T (0 < T < 1000) indicating the number of test cases. Then T test cases follow. Each
test case starts with an integer n (0 < n < 100). The ith line of the next n lines contains two integers tA, tB (0 < tA, tB < 100), giving the time to process the ith task by
machine A and machine B.

Output

For each test case, output the earliest time when all the tasks have been processed.

Sample Input

4
1
1 2
2
1 2
2 1
2
1 2
90 95
3
1 3
1 3
1 3

Sample Output

1
1
90
3

双塔DP

dp[i][j] 表示执行第i个任务，机器A和机器B的时间差

双塔DP 顾名思义可以把A机器和B机器看作是两座塔，如果把当前任务放在哪个塔上，那么哪个塔的高度就会增加

每次放任务，都会有两个选择，要么放在A塔，要么放在B塔。如果放在高的塔上，根据题意在同一个机器上必要上一个任务完成

时才能进行下一个任务，所以等高的塔完成上一个任务是，A，B都是空闲的了，这个是在高的塔上放任务，时间差就是这个任务的时间

如果放在低的塔上，那么在之前的差值上加上这个任务的时间

#include <iostream>
#include <string.h>
#include <stdlib.h>

#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 10000000
int dp[105][205];
int n;
int ta[105];
int tb[105];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d%d",&ta[i],&tb[i]);
for(int i=0;i<=n;i++)
for(int j=0;j<=200;j++)
dp[i][j]=MAX;
dp[0][0+100]=0;
for(int i=1;i<=n;i++)
{
for(int j=-99;j<=99;j++)
{
if(dp[i-1][j+100]==MAX)
continue;
if(j<0)
{
dp[i][-tb[i]+100]=min(dp[i][-tb[i]+100],dp[i-1][j+100]+tb[i]);
dp[i][j+ta[i]+100]=min(dp[i][j+ta[i]+100],dp[i-1][j+100]+max(0,j+ta[i]));
}
else
{
dp[i][ta[i]+100]=min(dp[i][ta[i]+100],dp[i-1][j+100]+ta[i]);
dp[i][j-tb[i]+100]=min(dp[i][j-tb[i]+100],dp[i-1][j+100]+max(0,tb[i]-j));
}
}
}
int ans=MAX;
for(int i=-99;i<=99;i++)
ans=min(ans,dp
[i+100]);
printf("%d\n",ans);
}
return 0;
}