【BZOJ 2748 [HAOI2012]音量调节】dp

最新推荐文章于 2021-04-13 22:18:29 发布

原创最新推荐文章于 2021-04-13 22:18:29 发布 · 178 阅读

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本文解析了一道名为BZOJ2748的问题，通过使用动态规划算法，具体为二维DP数组来记录每首歌可达的状态，最终求解出满足条件的最大值。代码中详细展示了从初始化到迭代更新的全过程。

BZOJ 2748
哇跟有一个gym一道题一模一样
可惜gym那题读沙雕了
中文题意
就是用dp[55][1025]
代表第几首歌可达的状态

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 1025;
int dp[55][MAX_N],arr[55];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,v,m;
    scanf("%d%d%d",&n,&v,&m);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    dp[0][v] = true;
    for(int i = 1;i<=n;++i)
    {
        for(int j = 0;j<=m;j++)
        {
            if(dp[i-1][j])
            {
                if(j+arr[i]<=m) dp[i][j+arr[i]] = 1;
                if(j-arr[i]>=0) dp[i][j-arr[i]] = 1;
            }
        }
    }
    int ans = -1;
    for(int i = m;i>=0;--i)
    {
        if(dp[n][i])
        {
            printf("%d\n",i);
            return 0;
        }
    }
    printf("%d\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}