Codeforces Round #306 (Div. 2) B. Preparing Olympiad dfs

题目链接:

http://codeforces.com/contest/550/problem/B

题意:

有n门课，然后让你选择一些课，要求这些课程的和小于等于r，大于等于l，最大值减去最小值至少为x
然后问你有多少种分法

题解:

数据范围小 n<=15 所以可以把n门课 变成二进制的状态
也可以直接dfs

代码:

dfs ~

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

int n,vis[20];
ll le,ri,x,a[20];
int ans;

void dfs(int t, ll sum, ll mi,ll mx){
    if(sum > ri) return ;

    if((mx-mi)>=x && (sum>=le) && (sum<=ri))
        ans++;
    for(int i=t; i<=n; i++){
        if(!vis[i]){
            vis[i] = 1;
            dfs(i,sum+a[i],min(a[i],mi),max(a[i],mx));
            vis[i] = 0;
        }
    }
}

int main(){
    n = read(),le=read(),ri=read(),x=read();
    for(int i=1; i<=n; i++)
        a[i] = read();

    dfs(1,0,INF,-1);

    cout << ans << endl;

    return 0;
}

每位表示状态 选和不选

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

ll n,le,ri,x,a[20];
int ans;

int main(){
    n = read(),le=read(),ri=read(),x=read();
    for(int i=0; i<n; i++)
        a[i] = read();

    for(int i=0; i<(1<<n); i++){
        ll cnt=0,sum=0,mi=INFLL,mx=-1;
        for(int j=0; j<n; j++){
            if(i&(1<<j)){
                cnt++;
                sum += a[j];
                mi = min(mi,a[j]);
                mx = max(mx,a[j]);
            }
        }
        if(cnt>=2 && (mx-mi)>=x && sum>=le && sum<=ri)
            ans++;
    }

    cout << ans << endl;

    return 0;
}