POJ 2570 線段樹

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Potted Flower

Time Limit: 2000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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DescriptionThe little cat takes over the management of a new park. There is a large circular statue in the center of the park, surrounded by N pots of flowers. Each potted flower will be assigned to an integer number (possibly negative) denoting how attractive it is. See the following graph as an example:

(Positions of potted flowers are assigned to index numbers in the range of 1 ... N. The i-th pot and the (i + 1)-th pot are consecutive for any given i (1 <= i < N), and 1st pot is next to N-th pot in addition.)



The board chairman informed the little cat to construct "ONE arc-style cane-chair" for tourists having a rest, and the sum of attractive values of the flowers beside the cane-chair should be as large as possible. You should notice that a cane-chair cannot be a total circle, so the number of flowers beside the cane-chair may be 1, 2, ..., N - 1, but cannot be N. In the above example, if we construct a cane-chair in the position of that red-dashed-arc, we will have the sum of 3+(-2)+1+2=4, which is the largest among all possible constructions.

Unluckily, some booted cats always make trouble for the little cat, by changing some potted flowers to others. The intelligence agency of little cat has caught up all the M instruments of booted cats‘ action. Each instrument is in the form of "A B", which means changing the A-th potted flowered with a new one whose attractive value equals to B. You have to report the new "maximal sum" after each instruction.InputThere will be a single test data in the input. You are given an integer N (4 <= N <= 100000) in the first input line.

The second line contains N integers, which are the initial attractive value of each potted flower. The i-th number is for the potted flower on the i-th position.

A single integer M (4 <= M <= 100000) in the third input line, and the following M lines each contains an instruction "A B" in the form described above.

Restriction: All the attractive values are within [-1000, 1000]. We guarantee the maximal sum will be always a positive integer.OutputFor each instruction, output a single line with the maximum sum of attractive values for the optimum cane-chair.Sample Input

53 -2 1 2 -542 -25 -52 -45 -1

Sample Output

4435

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define ls rt<<1#define rs rt<<1|1#define MAX(x,y) ((x)>(y)?(x):(y))#define MIN(x,y) ((x)>(y)?(y):(x))#define MAXN 100010int N,M;int a[MAXN];struct seg_tree{    int l,r;    int lmax,rmax,summax;    int lmin,rmin,summin;    int sum;}tree[MAXN*3];void PushUp(int rt){    tree[rt].sum=tree[ls].sum+tree[rs].sum;    tree[rt].lmax=MAX(tree[ls].lmax,tree[ls].sum+tree[rs].lmax);    tree[rt].rmax=MAX(tree[rs].rmax,tree[rs].sum+tree[ls].rmax);    tree[rt].summax=MAX(MAX(tree[ls].summax,tree[rs].summax),tree[ls].rmax+tree[rs].lmax);    tree[rt].lmin=MIN(tree[ls].lmin,tree[ls].sum+tree[rs].lmin);    tree[rt].rmin=MIN(tree[rs].rmin,tree[rs].sum+tree[ls].rmin);    tree[rt].summin=MIN(MIN(tree[ls].summin,tree[rs].summin),tree[ls].rmin+tree[rs].lmin);}void build(int rt,int l,int r){    if(l>r) return;    tree[rt].l=l;    tree[rt].r=r;    int mid=(l+r)/2;    if(l==r)    {        tree[rt].sum=a[l];        tree[rt].lmax=tree[rt].rmax=tree[rt].summax=a[l];        tree[rt].lmin=tree[rt].rmin=tree[rt].summin=a[l];        return;    }    build(rt*2,l,mid);    build(rt*2+1,mid+1,r);    PushUp(rt);}void update(int rt,int pos,int val){    if(tree[rt].l==tree[rt].r && tree[rt].l==pos)    {        tree[rt].sum=val;        tree[rt].lmax=tree[rt].rmax=tree[rt].summax=val;        tree[rt].lmin=tree[rt].rmin=tree[rt].summin=val;        return;    }    int m=(tree[rt].l + tree[rt].r)>>1;    if(pos<=m)        update(rt*2,pos,val);    else        update(rt*2+1,pos,val);    PushUp(rt);}int main(){    while(scanf("%d",&N)!=EOF)    {        for(int i=1;i<=N;i++)            scanf("%d",&a[i]);        build(1,1,N);        scanf("%d",&M);        int pos,val;        while(M--)        {            scanf("%d%d",&pos,&val);            update(1,pos,val);            if(tree[1].sum==tree[1].summax)                printf("%d\n",tree[1].sum-tree[1].summin);            else                printf("%d\n",MAX(tree[1].summax,tree[1].sum-tree[1].summin));        }    }    return 0;}

題意：給定一個環形序列，進行線上操作，每次修改一個元素，輸出環上的最大連續子列的和，但不能是完全序列。

分析:

      如果不是環的話，只是一個序列，用線段樹很方便求，所以將環從某一點切開成一個序列。

     那麼答案的最大連續和可能包含斷點（換種想法，包含斷點時的最大連續和即為 sum-區間最小的連續和）

    1，若是所有的數都大於0，那麼最大連續和（必須斷開一個）即為 總和sum-最小非空連續和。

    2，若是區間最小連續和小於0，那麼答案就是MAX(區間最大連續和，sum-最小非空連續和)。即分為包含斷點和不包含的比較。