165-C - Many Requirements

Time Limit: 2 sec / Memory Limit: 1024 MB

ps:一道dfs，我暂时还不会写，就看了一下。

Problem Statement

Given are positive integers $\mathrm{NN}$, $\mathrm{MM}$, $\mathrm{QQ}$, and $\mathrm{QQ}$ quadruples of integers ( ${\mathrm{aiai}}^{}$ , ${\mathrm{bibi}}^{}$ , ${\mathrm{cici}}^{}$ , ${\mathrm{didi}}^{}$ ).

Consider a sequence $\mathrm{AA}$ satisfying the following conditions:

• $\mathrm{AA}$ is a sequence of $\mathrm{NN}$ positive integers.
• $\mathrm{1\le A1\le A2\le \cdots \le AN\le M1\le A1\le A2\le \cdots \le AN\le M}$.

Let us define a score of this sequence as follows:

• The score is the sum of ${\mathrm{didi}}^{}$ over all indices $\mathrm{ii}$ such that ${\mathrm{Abi-Aai=ciAbi-Aai=ci}}^{}$. (If there is no such $\mathrm{ii}$, the score is $00$.)

Find the maximum possible score of $\mathrm{AA}$.

Constraints

• All values in input are integers.
• $\mathrm{2\le N\le 10}$
• $\mathrm{1\le M\le 10}$
• $\mathrm{1\le Q\le 50}$
• $\mathrm{1\le ai<bi\le N}$( $\mathrm{i=1,2,...,Q}$)
• $\mathrm{0\le ci\le M-1}$( $\mathrm{i=1,2,...,Q}$)
• $(ai,bi,ci)\ne (aj,bj,cj)$(where $\mathrm{i\ne j}$)
• $\mathrm{1\le di\le 105}$( $\mathrm{i=1,2,...,Q}$)

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Input

Input is given from Standard Input in the following format:

$N$ $M$ $Q$
${\mathrm{a1}}^{}$ ${\mathrm{b1}}^{}$ ${\mathrm{c1}}^{}$ ${\mathrm{d1}}^{}$
$::$
${\mathrm{aQ}}^{}$ ${\mathrm{bQ}}^{}$ ${\mathrm{cQ}}^{}$ ${\mathrm{dQ}}^{}$

Output

Print the maximum possible score of $A$.

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Sample Input 1 Copy

Copy

3 4 3
1 3 3 100
1 2 2 10
2 3 2 10

Sample Output 1 Copy

Copy

When $\mathrm{A=\{1,3,4\}}$, its score is $110$. Under these conditions, no sequence has a score greater than $110$, so the answer is $110$.

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Sample Input 2 Copy

Copy

4 6 10
2 4 1 86568
1 4 0 90629
2 3 0 90310
3 4 1 29211
3 4 3 78537
3 4 2 8580
1 2 1 96263
1 4 2 2156
1 2 0 94325
1 4 3 94328

Sample Output 2 Copy

Copy

357500

---

Sample Input 3 Copy

Copy

10 10 1
1 10 9 1

Sample Output 3 Copy

Copy

1

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
int N, M, Q, maxn = -9999999;
int a[101], b[101], c[101], d[101];
int flag[11], A[11];

void clear()
{
    int res = 0;
    for(int i=1;i<=Q;i++)
    {
        if(A[b[i]]-A[a[i]]==c[i])
            res += d[i];
    }
    if(res>maxn)
        maxn = res;
}

void dfs(int k)
{
    for(int i=A[k-1];i<=M;i++) //由于A递增 所以可以从前一个A的元素开始枚举
    {
        if(i!=0)
        {
            A[k] = i;
            if(k==N)
            {
                clear();
            }
            else dfs(k+1);
            A[k] = 0;
        }
    }
}

int main()
{
    ios::sync_with_stdio(0);
    scanf("%d%d%d", &N, &M, &Q);
    for(int i=1;i<=Q;i++)
    {
        scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
    }
    dfs(1);
    printf("%d\n", maxn);
    return 0;
}