CodeForces 550B Preparing Olympiad(DFS回溯+暴力枚举)
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【题目链接】:click here~~
【题目大意】
一组题目的数目(n<=15),每个题目有相应的难度,问你选择一定的题目(大于r个且小于l个)且选择后的题目里最小难度与最大难度差不小于x,求选择方案数。
【解题思路】:
DFS+回溯。
先发一发比较拙的代码:
#include <bits/stdc++.h>using namespace std;const int N=1e5+10;int num[N],mum[N];int n,m,q,t,l,r;int top,ans,cnt;void dfs(int sum,int d,int x,int minn)///和,极值,当前循环,最小数{ if(sum>=l&&sum<=r&&d>=q) ans++; if(sum>r) return; ///if(sum<l) return;当前累加的sum可能小于l for(int i=x;i<=n;++i) { dfs(sum+num[i],num[i]-minn,i+1,minn); }}int main(){ while(cin>>n>>l>>r>>q) { ans=0; memset(num,0,sizeof(num)); for(int i=1; i<=n; ++i) scanf("%d",&num[i]); sort(num+1,num+1+n); for(int i=1; i<=n; ++i) { dfs(num[i],0,i+1,num[i]);///为了参照最大和最小,引入最小数 } printf("%d\n",ans); } return 0;}
继续补充暴力枚举~~:
#include <bits/stdc++.h>using namespace std;const int N=1e2;int num[N];int n,l,r,x;int main(){ scanf("%d%d%d%d",&n,&l,&r,&x); int ans=0; memset(num,0,sizeof(num)); for(int i=0; i<n; ++i) scanf("%d",&num[i]); for(int s=0; s<(1<<n); ++s) { int cnt=0,sum=0; int maxx=-1e8,minn=1e8; for(int i=0; i<n; ++i) if((s>>i)&1)///关键! { sum+=num[i]; maxx=max(num[i],maxx); minn=min(num[i],minn); cnt++; } if(cnt>=2&&maxx-minn>=x&&sum>=l&&sum<=r) ans++; } printf("%d\n",ans); return 0;}
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