POJ 2750 Potted Flower 线段树+DP

题意：给定一个环形的序列，值可正可负，求最大的连续子序列。（每次更新都需要输出结果，最多包含N-1个点）。

#include <iostream>using namespace std;#define N 100005int pot[N];struct item{int sum;int lmax,lmin;int rmax,rmin;int submax, submin;int left, right;} node[N*6];int max ( int a, int b ){return ( a > b ? a : b );}int min ( int a, int b ){return ( a < b ? a : b );}void update ( int root ){int lchild = root * 2;int rchild = root * 2 + 1;node[root].sum = node[lchild].sum + node[rchild].sum;  node[root].lmax = max ( node[lchild].lmax, node[lchild].sum + node[rchild].lmax ); node[root].lmin = min ( node[lchild].lmin, node[lchild].sum + node[rchild].lmin );  node[root].rmax = max ( node[rchild].rmax, node[lchild].rmax + node[rchild].sum );node[root].rmin = min ( node[rchild].rmin, node[lchild].rmin + node[rchild].sum );  node[root].submax = max ( max ( node[lchild].submax, node[rchild].submax ), node[lchild].rmax + node[rchild].lmax );  node[root].submin = min ( min ( node[lchild].submin, node[rchild].submin ), node[lchild].rmin + node[rchild].lmin ); }void build_tree ( int l, int r, int root ){node[root].left = l;node[root].right = r;if ( l == r ){node[root].sum = node[root].lmax = node[root].lmin = node[root].rmax = pot[l];node[root].rmin = node[root].submax = node[root].submin = pot[l];return;}int mid = ( l + r ) / 2;build_tree ( l, mid, root * 2 );build_tree ( mid + 1, r, root * 2 + 1 );update ( root );}void modify ( int pos, int val, int root ){if ( node[root].left == node[root].right && node[root].left == pos ){node[root].sum = node[root].lmax = node[root].lmin = node[root].rmax = val;node[root].rmin = node[root].submax = node[root].submin = val;return;}int mid = ( node[root].left + node[root].right ) / 2;if ( pos <= mid )modify ( pos, val, root * 2 );elsemodify ( pos, val, root * 2 + 1 );update ( root );}int main(){int n, m, pos, val, i;scanf("%d",&n);for ( i = 1; i <= n; ++i )scanf("%d",pot+i);build_tree ( 1, n, 1 );scanf("%d",&m);while ( m-- ){scanf("%d%d",&pos,&val);modify(pos,val,1);if ( node[1].sum == node[1].submax )printf("%d\n",node[1].sum - node[1].submin);elseprintf("%d\n", max ( node[1].sum - node[1].submin, node[1].submax ) );}return 0;}