HDU 4381 Grid【0-1背包】

The story happened long long ago. One day, Cao Cao made a special order called “Chicken Rib” to his army. No one got his point and all became very panic. However, Cao Cao himself felt very proud of his interesting idea and enjoyed it.

Xiu Yang, one of the cleverest counselors of Cao Cao, understood the command Rather than keep it to himself, he told the point to the whole army. Cao Cao got very angry at his cleverness and would like to punish Xiu Yang. But how can you punish someone because he’s clever? By looking at the chicken rib, he finally got a new idea to punish Xiu Yang.

He told Xiu Yang that as his reward of encrypting the special order, he could take as many gold sticks as possible from his desk. But he could only use one stick as the container.

Formally, we can treat the container stick as an LL length segment. And the gold sticks as segments too. There were many gold sticks with different length aiai and value vivi. Xiu Yang needed to put these gold segments onto the container segment. No gold segment was allowed to be overlapped. Luckily, Xiu Yang came up with a good idea. On the two sides of the container, he could make part of the gold sticks outside the container as long as the center of the gravity of each gold stick was still within the container. This could help him get more valuable gold sticks.

As a result, Xiu Yang took too many gold sticks which made Cao Cao much more angry. Cao Cao killed Xiu Yang before he made himself home. So no one knows how many gold sticks Xiu Yang made it in the container.

Can you help solve the mystery by finding out what’s the maximum value of the gold sticks Xiu Yang could have taken?
Input
The first line of the input gives the number of test cases, T(1≤T≤100)T(1≤T≤100). TT test cases follow. Each test case start with two integers, N(1≤N≤1000)N(1≤N≤1000) and L(1≤L≤2000)L(1≤L≤2000), represents the number of gold sticks and the length of the container stick. NN lines follow. Each line consist of two integers, ai(1≤ai≤2000)ai(1≤ai≤2000) and vi(1≤vi≤109)vi(1≤vi≤109), represents the length and the value of the ithith gold stick.
Output
For each test case, output one line containing Case #x: y, where xx is the test case number (starting from 1) and yy is the maximum value of the gold sticks Xiu Yang could have taken.
Sample Input
4

3 7
4 1
2 1
8 1

3 7
4 2
2 1
8 4

3 5
4 1
2 2
8 9

1 1
10 3
Sample Output
Case #1: 2
Case #2: 6
Case #3: 11
Case #4: 3

Hint
In the third case, assume the container is lay on x-axis from 0 to 5. Xiu Yang could put the second gold stick center at 0 and put the third gold stick center at 5,
so none of them will drop and he can get total 2+9=11 value.

In the fourth case, Xiu Yang could just put the only gold stick center on any position of [0,1], and he can get the value of 3.

题意：给出一个序列，序列上所有的点都被标记为黑色。
下面有两种操作，第一种操作是，使0~ai区间内x个元素变成白色，第二种操作是在ai~n区间内x个元素变成白色。
求变成白色最多状态下的最小花费。

分析：
每种操作只限用一次，这是一道变形的0-1背包问题。显然最后需要时两份花费取最小值，所以一开始我们通过两个部分来跑背包。后面的状态可以处理一下，也就可以变成从0开始的背包问题。

最后需要枚举 dp[i]+dp[j] 的两次背包的状态，维护最优解。

对于DP的理解还是不够深刻！

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn = 1e3 + 5;
struct node {
    int x, num;
    bool operator <(const node &b) const
    {
        return x < b.x;
    }
    /*bool_constant operator <(node &b) 
    {
        return x < b.x;
    }*/
}p1[maxn],p2[maxn];
int dp1[maxn], dp2[maxn];
int main()
{
    int T; int cs = 0;
    scanf("%d", &T);
    while (T--)
    {
        cs++;
        int n, m;
        scanf("%d%d", &n, &m);
        memset(dp1, INF, sizeof dp1);
        memset(dp2, INF, sizeof dp2);
        int k1 = 0, k2 = 0;
        for (int i = 0; i<m; i++)
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            if (u == 1)
            {
                p1[k1].x = v;
                p1[k1++].num = w;
            }
            else
            {
                p2[k2].x = n + 1 - v;//处理成 可以从0开始DP的状态
                p2[k2++].num = w;
            }
        }
        sort(p1, p1 + k1);
        sort(p2, p2 + k2);
        dp1[0] = dp2[0] = 0;
        //存在，值才正常
        for (int i = 0; i<k1; i++)
        {
            for (int j = p1[i].x; j >= p1[i].num; j--)
            {
                dp1[j] = min(dp1[j], dp1[j - p1[i].num] + 1);
            }
        }
        for (int i = 0; i<k2; i++)
        {
            for (int j = p2[i].x; j >= p2[i].num; j--)
            {
                dp2[j] = min(dp2[j], dp2[j - p2[i].num] + 1);
            }
        }
        //枚举涂的个数
        int tmp, ans = 0, sum = 0;
        for (int i = 1; i <= n; i++)
        {
            for (int j = 0; j <= i; j++)
            {
                tmp = dp1[j] + dp2[i - j];
                if (tmp <= m)
                {
                    if (ans != i)
                    {
                        ans = i;
                        sum = tmp;
                    }
                    else
                    {
                        sum = min(sum, tmp);
                    }
                }
            }
        }
        printf("Case %d: %d %d\n", cs, ans, sum);
    }
    return 0;
}