LA 3938 动态最大连续和 线段树
题目链接：
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1939
来自：刘汝佳大白书P201.
解题思路：
构造一棵线段树，其中每个结点维护三个值，记录最大前缀和，最大后缀和最大连续和。
最大连续和要么完全在左段，要么完全在右段，要么在跨越中线。就是会是左段的最大后缀和+右段的最大前缀和。。。。
代码也是刘汝佳写的哦
代码：
1 // LA3938 Ray, Pass me the dishes! 2 // Rujia Liu 3 #include<cstdio> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7 8 const int maxn = 500000 + 10; 9 const int maxnode = 1000000 + 10; 10 typedef long long LL; 11 typedef pair<int,int> Interval; 12 13 LL prefix_sum[maxn]; 14 15 LL sum(int L, int R) 16 { 17 return prefix_sum[R] - prefix_sum[L-1]; 18 } 19 20 LL sum(Interval p) 21 { 22 return sum(p.first, p.second); 23 } 24 25 Interval better(Interval a, Interval b) 26 { 27 if(sum(a) != sum(b)) return sum(a) > sum(b) ? a : b; 28 return a < b ? a : b; // 利用pair自带的字典序 29 } 30 31 int qL, qR; 32 33 struct IntervalTree 34 { 35 int max_prefix[maxnode]; 36 int max_suffix[maxnode]; 37 Interval max_sub[maxnode]; 38 39 void build(int o, int L, int R) 40 { 41 if(L == R) 42 { 43 max_prefix[o] = max_suffix[o] = L; 44 max_sub[o] = make_pair(L, L); 45 } 46 else 47 { 48 int M = L + (R-L)/2; 49 // 递归创建子树 50 int lc = o*2, rc = o*2+1; 51 build(lc, L, M); 52 build(rc, M+1, R); 53 54 // 递推max_prefix 55 LL v1 = sum(L, max_prefix[lc]); 56 LL v2 = sum(L, max_prefix[rc]); 57 if(v1 == v2) max_prefix[o] = min(max_prefix[lc], max_prefix[rc]); 58 else max_prefix[o] = v1 > v2 ? max_prefix[lc] : max_prefix[rc]; 59 60 // 递推max_suffix 61 v1 = sum(max_suffix[lc], R); 62 v2 = sum(max_suffix[rc], R); 63 if(v1 == v2) max_suffix[o] = min(max_suffix[lc], max_suffix[rc]); 64 else max_suffix[o] = v1 > v2 ? max_suffix[lc] : max_suffix[rc]; 65 66 // 递推max_sub 67 max_sub[o] = better(max_sub[lc], max_sub[rc]); // 完全在左子树或者右子树 68 max_sub[o] = better(max_sub[o], make_pair(max_suffix[lc], max_prefix[rc])); // 跨越中线 69 } 70 } 71 72 Interval query_prefix(int o, int L, int R) 73 { 74 if(max_prefix[o] <= qR) return make_pair(L, max_prefix[o]); 75 int M = L + (R-L)/2; 76 int lc = o*2, rc = o*2+1; 77 if(qR <= M) return query_prefix(lc, L, M); 78 Interval i = query_prefix(rc, M+1, R); 79 i.first = L; 80 return better(i, make_pair(L, max_prefix[lc])); 81 } 82 83 Interval query_suffix(int o, int L, int R) 84 { 85 if(max_suffix[o] >= qL) return make_pair(max_suffix[o], R); 86 int M = L + (R-L)/2; 87 int lc = o*2, rc = o*2+1; 88 if(qL > M) return query_suffix(rc, M+1, R); 89 Interval i = query_suffix(lc, L, M); 90 i.second = R; 91 return better(i, make_pair(max_suffix[rc], R)); 92 } 93 94 Interval query(int o, int L, int R) 95 { 96 if(qL <= L && R <= qR) return max_sub[o]; 97 int M = L + (R-L)/2; 98 int lc = o*2, rc = o*2+1; 99 if(qR <= M) return query(lc, L, M); 100 if(qL > M) return query(rc, M+1, R); 101 Interval i1 = query_prefix(rc, M+1, R); // 右半的前缀 102 Interval i2 = query_suffix(lc, L, M); // 左半的后缀 103 Interval i3 = better(query(lc, L, M), query(rc, M+1, R)); 104 return better(make_pair(i2.first, i1.second), i3); 105 } 106 }; 107 108 IntervalTree tree; 109 110 int main() 111 { 112 int kase = 0, n, a, Q; 113 while(scanf("%d%d", &n, &Q) == 2) 114 { 115 prefix_sum[0] = 0; 116 for(int i = 0; i < n; i++) 117 { 118 scanf("%d", &a); 119 prefix_sum[i+1] = prefix_sum[i] + a; 120 } 121 tree.build(1, 1, n); 122 printf("Case %d:\n", ++kase); 123 while(Q--) 124 { 125 int L, R; 126 scanf("%d%d", &L, &R); 127 qL = L; 128 qR = R; 129 Interval ans = tree.query(1, 1, n); 130 printf("%d %d\n", ans.first, ans.second); 131 } 132 } 133 return 0; 134 }
