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题意：给一个array，有两种操作，(1)修改某一个位置的值，(2)询问区间[L,R]内的最大子段和，其中子段需满足相邻两个数的位置的奇偶性不同
思路：假设对于询问操作没有奇偶性的限制，那么记录区间的最大子段和就可以通过合并区间得到答案了。加上奇偶性的限制后，记录的信息必须更加具体，需要把子段的端点的奇偶性加进去，也就是说一个区间需要记录4个值， 分别是奇奇，奇偶，偶偶，偶奇，然后同样可以通过合并区间来得到答案。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define foreach(i, a) for (typeof(a.begin()) it = a.begin(); it != a.end(); it ++) // //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // ///* -------------------------------------------------------------------------------- */ //template<typename T>bool umax(T &a, const T &b) { return a >= b? false : (a = b, true);}#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1const ll inf = (ll)1e18;const int maxn = 1e5 + 7;struct SegTree {private: struct Node { ll a[4]; }; Node tree[maxn << 2]; int n; bool chk(int i, int j) { return (i & 1) ^ (j >> 1); } int get(int i, int j) { return (i & 2) | (j & 1); } Node merge(const Node &nl, const Node &nr) { Node ans; for (int i = 0; i < 4; i ++) { ans.a[i] = nl.a[i]; umax(ans.a[i], nr.a[i]); } for (int i = 0; i < 4; i ++) { for (int j = 0; j < 4; j ++) { if (chk(i, j)) { umax(ans.a[get(i, j)], nl.a[i] + nr.a[j]); } } } return ans; } void build(int l, int r, int rt) { if (l == r) { int x; RI(x); int buf = (l & 1) << 1 | (l & 1); for (int i = 0; i < 4; i ++) tree[rt].a[i] = i == buf? x : -inf; return ; } int m = (l + r) >> 1; build(lson); build(rson); tree[rt] = merge(tree[rt << 1], tree[rt << 1 | 1]); } void update(int p, int x, int l, int r, int rt) { if (l == r) { tree[rt].a[(p & 1) << 1 | (p & 1)] = x; return ; } int m = (l + r) >> 1; if (p <= m) update(p, x, lson); else update(p, x, rson); tree[rt] = merge(tree[rt << 1], tree[rt << 1 | 1]); } Node query(int L, int R, int l, int r, int rt) { if (L <= l && r <= R) return tree[rt]; int m = (l + r) >> 1; if (R <= m) return query(L, R, lson); if (L > m) return query(L, R, rson); return merge(query(L, R, lson), query(L, R, rson)); }public: void build(int nn) { n = nn; build(1, n, 1); } void update(int p, int x) { update(p, x, 1, n, 1); } ll query(int l, int r) { Node buf = query(l, r, 1, n, 1); ll ans = buf.a[0]; for (int i = 1; i < 4; i ++) umax(ans, buf.a[i]); return ans; }};SegTree st;int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE int T; cin >> T; while (T --) { int n, m; RI(n, m); st.build(n); for (int i = 0; i < m; i ++) { int t, a, b; RI(t, a, b); if (t) st.update(a, b); else printf("%I64d\n", st.query(a, b)); } } return 0; //} // // // ///* ******************************************************************************** */ |
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原文地址：http://www.cnblogs.com/jklongint/p/4684138.html
