一晚上只做一道题——细节我给跪

周三，还是道求最大子列和的问题，一个字母。。。跪了

HDU 1003 Max Sum

Given a sequence a1,a2,a3……an
, your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

是时候给最大子列和的问题完结一下了；这是一类动态规划的经典题目，但是我却喜欢用另外一种方法，因为这类问题很特殊，如果不要求最大子列的起始位置与结束的位置，那么有一种生气的方法叫做在线处理（代码如下）

这个代码复杂度是o(n)，接近是一个完美的算法了，只要遍历一遍就可以求出答案。

那么还可以用分治的做法，求出将问题分解成小规模问题后，求出左边的最大子列，再求出右边的最大子列，最后求出跨分界线的最大子列，求三个中的max即可；代码如下：

#include <iostream>#include <cstdlib>using namespace std;int List[10050];int l[10], r[10];int max_num(int a, int b, int c){//求三个数最大值    int Max;    if(a>b)Max=a;    else Max=b;    if(Max>c)return Max;    else return c;}int divide(int List[], int l, int r){    int Lsum, Rsum;    int Leftpart, Rightpart;    int Lmaxsum,Rmaxsum;    int mid, i;    if(l==r){        if(List[l]>0)return List[l];        else return 0;    }    mid=(r-l)/2+l;    Lsum=divide(List,l,mid);    Rsum=divide(List,mid+1,r);    Lmaxsum=0;Leftpart=0;    for(i=mid;i>=l;i--){        Leftpart+=List[i];        if(Leftpart>Lmaxsum)Lmaxsum=Leftpart;    }    Rmaxsum=0;Rightpart=0;    for(i=mid+1;i<=r;i++){        Rightpart+=List[i];        if(Rightpart>Rmaxsum)Rmaxsum=Rightpart;    }    return max_num(Lsum,Rsum,Lmaxsum+Rmaxsum);}int max_list(int List[],  int n){    return divide(List, 0, n-1);}int main(){    int n;    while(cin>>n&&n){        int i, j=0, tmp=0, maxx=-1, st=0, ed=n-1;        for(i=0;i<n;i++)            cin>>List[i];        for(i=0;i<n;i++){            tmp+=List[i];            if(tmp>maxx){                 st=j;ed=i;maxx=tmp;            }            else if(tmp<0){                j=i+1;                tmp=0;            }        }        cout<<max_list(List, n)<<" "<<List[st]<<" "<<List[ed]<<endl;    }    return 0;}

动态规划的想法也很好，状态转移方程很简单 如下，附代码：

 if((dp[i-1]+a[i])>=a[i]){            dp[i]=dp[i-1]+a[i];            st[i]=st[i-1];            last[i]=i;        }else {            dp[i]=a[i];            st[i]=last[i]=i;        }

#include <iostream>#include <algorithm>#define inf 0x3f3f3f3fusing namespace std;int main(){    int k,res,i,t=0,n,cnt=1;    cin>>n;    while(n--){    cin>>k;    int a[123456]={0},dp[123456]={0},st[123456]={0},last[123456]={0};        res=-inf;        for(i=0;i<k;i++){            cin>>a[i];        }    dp[0]=a[0];    st[0]=last[0]=0;    for(i=1;i<k;i++){        if((dp[i-1]+a[i])>=a[i]){            dp[i]=dp[i-1]+a[i];            st[i]=st[i-1];            last[i]=i;        }else {            dp[i]=a[i];            st[i]=last[i]=i;        }    }    for(i=0;i<k;i++){        if(res<dp[i]){            t=i;            res=dp[i];        }    }    cout<<"Case "<<cnt<<":"<<endl;    cout<<res<<" "<<st[t]+1<<" "<<last[t]+1<<endl;    if(n)    cout<<endl;    cnt++;    }    return 0;}

我最喜欢的方法，注意点是maxx赋值可以给一个很小的值inf=-0x3f3f3f3f，或者是数据的下限减一；然后数据初始化要放对位置，放在while（n–）{…}里面，遇到temp<0时，原来的尾巴下标+1就是新的头，即j=i+1;
tips：数组第一位a[0]可以不用，直接i=1起步；代码如下：

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[100010];int main(){    int t,p;    cin>>t;    p=t;    while(t--){        int n,i;        cin>>n;        int st=1,ed=1,j=1,temp=0,maxx=-1001;        for(i=1;i<=n;i++)            cin>>a[i];        for(i=1;i<=n;i++){            temp+=a[i];            if(temp>maxx){                maxx=temp;st=j;ed=i;            }            if(temp<0){                temp=0;j=i+1;            }        }        cout<<"Case "<<p-t<<":"<<endl;        cout<<maxx<<" "<<st<<" "<<ed<<endl;        if(t)cout<<endl;        }    return 0;}

哈哈哈，今天到此结束，上铂金啦，困高困高。。