hdu 1231 最大连续子序列 yy+dp+数据结构解法

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我是靠谱客的博主无情天空，最近开发中收集的这篇文章主要介绍hdu 1231 最大连续子序列 yy+dp+数据结构解法，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

1:线段树解法对于每一个i求出最大的（i-k）区间内的sum值及id

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<sstream>
#include<string>
#include<climits>
#include<stack>
#include<set>
#include<bitset>
#include<cmath>
#include<deque>
#include<map>
#include<queue>
#define iinf 0x7f7f7f7f
#define linf 1000000000000000000LL
#define dinf 1e200
#define eps 1e-11
#define all(v) (v).begin(),(v).end()
#define sz(x)
x.size()
#define pb push_back
#define mp make_pair
#define lng long long
#define sqr(a) ((a)*(a))
#define pii pair<int,int>
#define pll pair<lng,lng>
#define pss pair<string,string>
#define pdd pair<double,double>
#define X first
#define Y second
#define pi 3.14159265359
#define ff(i,xi,n) for(int i=xi;i<=(int)(n);++i)
#define ffd(i,xi,n) for(int i=xi;i>=(int)(n);--i)
#define ffl(i,r) for(int i=head[r];i!=-1;i=edge[i].next)
#define cc(i,j) memset(i,j,sizeof(i))
#define two(x)
((lng)1<<(x))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mod
1000000007
#define pmod(x,y) (x%y+y)%y
using namespace std;
typedef vector<int>
vi;
typedef vector<string>
vs;
template<class T> inline void checkmax(T &x,T y){if(x<y) x=y;}
template<class T> inline void checkmin(T &x,T y){if(x>y) x=y;}
template<class T> inline T Min(T x,T y){return (x>y?y:x);}
template<class T> inline T Max(T x,T y){return (x<y?y:x);}
template<class T> T Abs(T a){return a>0?a:(-a);}
template<class T> inline T lowbit(T n){return (n^(n-1))&n;}
template<class T> inline int countbit(T n){return (n==0)?0:(1+countbit(n&(n-1)));}
int k,a[11111],sum[11111],s[44444];
void push_up(int rt)
{
if(sum[s[rt<<1]]>=sum[s[rt<<1|1]])
s[rt]=s[rt<<1];
else
s[rt]=s[rt<<1|1];
}
void build(int l,int r,int rt)
{
if(l==r)
{
s[rt]=l;
return ;
}
int mid=(l+r)>>1;
build(lson);
build(rson);
push_up(rt);
}
int query(int L,int R,int l,int r,int rt)
{
if(l>=L&&R>=r)
{
return s[rt];
}
int ret=-iinf;
int mid=(l+r)>>1;
if(L<=mid)
{
ret=query(L,R,lson);
}
if(R>mid)
{
int y;
if(ret==-iinf)
ret=query(L,R,rson);
else
if(sum[ret]<sum[y=query(L,R,rson)])
ret=y;
}
return ret;
}
int main()
{
while(scanf("%d",&k)==1)
{
if(k==0) break;
ff(i,1,k) scanf("%d",a+i);
sum[0]=0;
ff(i,1,k) sum[i]=sum[i-1]+a[i];
int ans1=-iinf,le,ri;
build(1,k,1);
ff(i,1,k)
{
int r;
int u=sum[r=query(i,k,1,k,1)]-sum[i-1];
//
cout<<u<<" "<<r<<endl;
if(u>ans1) ans1=u,le=i,ri=r;
//
cout<<ans1<<" "<<le<<" "<<ri<<endl;
}
if(ans1<0) ans1=0,le=1,ri=k;
printf("%d %d %dn",ans1,a[le],a[ri]);
}
return 0;
}

2： YY解法贪心

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<sstream>
#include<string>
#include<climits>
#include<stack>
#include<set>
#include<bitset>
#include<cmath>
#include<deque>
#include<map>
#include<queue>
#define iinf 0x7f7f7f7f
#define linf 1000000000000000000LL
#define dinf 1e200
#define eps 1e-11
#define all(v) (v).begin(),(v).end()
#define sz(x)
x.size()
#define pb push_back
#define mp make_pair
#define lng long long
#define sqr(a) ((a)*(a))
#define pii pair<int,int>
#define pll pair<lng,lng>
#define pss pair<string,string>
#define pdd pair<double,double>
#define X first
#define Y second
#define pi 3.14159265359
#define ff(i,xi,n) for(int i=xi;i<=(int)(n);++i)
#define ffd(i,xi,n) for(int i=xi;i>=(int)(n);--i)
#define ffl(i,r) for(int i=head[r];i!=-1;i=edge[i].next)
#define cc(i,j) memset(i,j,sizeof(i))
#define two(x)
((lng)1<<(x))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mod
1000000007
#define pmod(x,y) (x%y+y)%y
using namespace std;
typedef vector<int>
vi;
typedef vector<string>
vs;
template<class T> inline void checkmax(T &x,T y){if(x<y) x=y;}
template<class T> inline void checkmin(T &x,T y){if(x>y) x=y;}
template<class T> inline T Min(T x,T y){return (x>y?y:x);}
template<class T> inline T Max(T x,T y){return (x<y?y:x);}
template<class T> T Abs(T a){return a>0?a:(-a);}
template<class T> inline T lowbit(T n){return (n^(n-1))&n;}
template<class T> inline int countbit(T n){return (n==0)?0:(1+countbit(n&(n-1)));}
int k,a[11111];
int main()
{
while(scanf("%d",&k)==1)
{
if(k==0) break;
ff(i,1,k) scanf("%d",a+i);
int sum=0;
int l=1,r,ans=-iinf,le,ri;
ff(i,1,k)
{
if(sum+a[i]<0)
{
sum+=a[i];
if(ans<sum)
ans=sum,le=l,ri=r;
sum=0;
l=i+1;
}
else
{
r=i;
sum+=a[i];
if(ans<sum)
ans=sum,le=l,ri=r;
}
}
if(ans<0) ans=0,le=1,ri=k;
printf("%d %d %dn",ans,a[le],a[ri]);
}
return 0;
}

3：dp解法

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<sstream>
#include<string>
#include<climits>
#include<stack>
#include<set>
#include<bitset>
#include<cmath>
#include<deque>
#include<map>
#include<queue>
#define iinf 0x7f7f7f7f
#define linf 1000000000000000000LL
#define dinf 1e200
#define eps 1e-11
#define all(v) (v).begin(),(v).end()
#define sz(x)
x.size()
#define pb push_back
#define mp make_pair
#define lng long long
#define sqr(a) ((a)*(a))
#define pii pair<int,int>
#define pll pair<lng,lng>
#define pss pair<string,string>
#define pdd pair<double,double>
#define X first
#define Y second
#define pi 3.14159265359
#define ff(i,xi,n) for(int i=xi;i<=(int)(n);++i)
#define ffd(i,xi,n) for(int i=xi;i>=(int)(n);--i)
#define ffl(i,r) for(int i=head[r];i!=-1;i=edge[i].next)
#define cc(i,j) memset(i,j,sizeof(i))
#define two(x)
((lng)1<<(x))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mod
1000000007
#define pmod(x,y) (x%y+y)%y
using namespace std;
typedef vector<int>
vi;
typedef vector<string>
vs;
template<class T> inline void checkmax(T &x,T y){if(x<y) x=y;}
template<class T> inline void checkmin(T &x,T y){if(x>y) x=y;}
template<class T> inline T Min(T x,T y){return (x>y?y:x);}
template<class T> inline T Max(T x,T y){return (x<y?y:x);}
template<class T> T Abs(T a){return a>0?a:(-a);}
template<class T> inline T lowbit(T n){return (n^(n-1))&n;}
template<class T> inline int countbit(T n){return (n==0)?0:(1+countbit(n&(n-1)));}
int dp[11111],k,le[11111],a[11111];
int main()
{
while(scanf("%d",&k)==1)
{
if(!k) break;
ff(i,1,k) scanf("%d",a+i);
dp[0]=-iinf;
ff(i,1,k)
if(dp[i-1]+a[i]<a[i]) dp[i]=a[i],le[i]=i;
else
dp[i]=dp[i-1]+a[i],le[i]=le[i-1];
int ans=-iinf,l,r;
ff(i,1,k) if(ans<dp[i]) ans=dp[i],l=le[i],r=i;
if(ans<0) ans=0,l=1,r=k;
printf("%d %d %dn",ans,a[l],a[r]);
}
return 0;
}

总的来看： yy思路以及线段树思路来的比较容易 (当然st算法也可以做·我就不写了）