poj 2750 Potted Flower(线段树区间合并)

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http://poj.org/problem?id=2750

有ｎ个数围成一个圈，每次可以将ａ位置上的数变为ｂ，对每个操作，输出区间的最大连续子段和，连续的子段长度不能超过ｎ。

区间合并问题，因为是求连续子段的和。先把圈从１和ｎ之间断开，变为一条链，先在链上求最长连续的和。这个最长连续的和取左节点最长连续和，右节点最长连续和，左节点从右边数最大连续和加上右节点从左边数最大连续和三者的最大值。

特殊情况就是当区间全为正的时候，最长连续和等于１～ｎ的和，这违背题意，它应该等于区间总和减去区间内最小连续的和。因此节点内在记录最大连续和的同时还要记录最小连续和。剩下的就是push_up和单点更新了。

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL __int64#define eps 1e-12#define PI acos(-1.0)#define PP pair<LL,LL>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 100010;const int mod = 1000000007;struct node{int l,r;int lmin,rmin;//分别从左右两边数最小连续的和int lmax,rmax;//分别从左右两边数最大连续的和int sum,smax,smin;//区间总和，最大连续和，最小连续和}tree[maxn*4];int a[maxn];void push_up(int v){tree[v].sum = tree[v*2].sum + tree[v*2+1].sum;tree[v].lmax = max(tree[v*2].lmax, tree[v*2].sum + tree[v*2+1].lmax);tree[v].lmin = min(tree[v*2].lmin, tree[v*2].sum + tree[v*2+1].lmin);tree[v].rmax = max(tree[v*2+1].rmax, tree[v*2+1].sum + tree[v*2].rmax);tree[v].rmin = min(tree[v*2+1].rmin, tree[v*2+1].sum + tree[v*2].rmin);tree[v].smax = max(max(tree[v*2].smax,tree[v*2+1].smax),tree[v*2].rmax+tree[v*2+1].lmax);tree[v].smin = min(min(tree[v*2].smin,tree[v*2+1].smin),tree[v*2].rmin+tree[v*2+1].lmin);}void build(int v, int l, int r){tree[v].l = l;tree[v].r = r;if(l == r){tree[v].lmax = tree[v].lmin = tree[v].rmax = tree[v].rmin = a[l];tree[v].smax = tree[v].smin = tree[v].sum = a[l];return;}int mid = (l+r) >> 1;build(v*2,l,mid);build(v*2+1,mid+1,r);push_up(v);}void update(int v, int pos, int val){if(tree[v].l == tree[v].r){tree[v].lmax = tree[v].lmin = tree[v].rmax = tree[v].rmin = val;tree[v].smax = tree[v].smin = tree[v].sum = val;return;}int mid = (tree[v].l + tree[v].r) >> 1;if(pos <= mid)update(v*2,pos,val);else update(v*2+1,pos,val);push_up(v);}int main(){int n,m;int x,y;while(~scanf("%d",&n)){for(int i = 1; i <= n; i++)scanf("%d",&a[i]);scanf("%d",&m);build(1,1,n);for(int i = 1; i <= m; i++){scanf("%d %d",&x,&y);update(1,x,y);if(tree[1].smax == tree[1].sum && tree[1].sum > 0)printf("%d\n",tree[1].sum - tree[1].smin);else{printf("%d\n",max(tree[1].smax,tree[1].sum - tree[1].smin));}}}return 0;}

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