题目链接：https://atcoder.jp/contests/abc140/tasks/abc140_e
给定一个 1~N 的排列 P.
定义$X_{L, R}$的值为$P_L, P_{L+1}, \ldots, P_R$中第二大的数的值.
求$\displaystyle \sum_{L=1}^{N-1} \sum_{R=L+1}^{N} X_{L,R}$.
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
12
13 #define pr(x) cout << #x << " = " << x << " "
14 #define prln(x) cout << #x << " = " << x << endl
15
16 #define LOWBIT(x) ((x)&(-x))
17
18 #define ALL(x) x.begin(),x.end()
19 #define INS(x) inserter(x,x.begin())
20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // ?? x ?????? c
22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
24
25 #define ms0(a) memset(a,0,sizeof(a))
26 #define msI(a) memset(a,inf,sizeof(a))
27 #define msM(a) memset(a,-1,sizeof(a))
28
29 #define MP make_pair
30 #define PB push_back
31 #define ft first
32 #define sd second
33
34 template<typename T1, typename T2>
35 istream &operator>>(istream &in, pair<T1, T2> &p) {
36 in >> p.first >> p.second;
37 return in;
38 }
39
40 template<typename T>
41 istream &operator>>(istream &in, vector<T> &v) {
42 for (auto &x: v)
43 in >> x;
44 return in;
45 }
46
47 template<typename T>
48 ostream &operator<<(ostream &out, vector<T> &v) {
49 Rep(i, v.size()) out << v[i] << " \n"[i == v.size()];
50 return out;
51 }
52
53 template<typename T1, typename T2>
54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
55 out << "[" << p.first << ", " << p.second << "]" << "\n";
56 return out;
57 }
58
59 inline int gc(){
60 static const int BUF = 1e7;
61 static char buf[BUF], *bg = buf + BUF, *ed = bg;
62
63 if(bg == ed) fread(bg = buf, 1, BUF, stdin);
64 return *bg++;
65 }
66
67 inline int ri(){
68 int x = 0, f = 1, c = gc();
69 for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
70 for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
71 return x*f;
72 }
73
74 template<class T>
75 inline string toString(T x) {
76 ostringstream sout;
77 sout << x;
78 return sout.str();
79 }
80
81 inline int toInt(string s) {
82 int v;
83 istringstream sin(s);
84 sin >> v;
85 return v;
86 }
87
88 //min <= aim <= max
89 template<typename T>
90 inline bool BETWEEN(const T aim, const T min, const T max) {
91 return min <= aim && aim <= max;
92 }
93
94 typedef long long LL;
95 typedef unsigned long long uLL;
96 typedef pair< double, double > PDD;
97 typedef pair< int, int > PII;
98 typedef pair< int, PII > PIPII;
99 typedef pair< string, int > PSI;
100 typedef pair< int, PSI > PIPSI;
101 typedef set< int > SI;
102 typedef set< PII > SPII;
103 typedef vector< int > VI;
104 typedef vector< double > VD;
105 typedef vector< VI > VVI;
106 typedef vector< SI > VSI;
107 typedef vector< PII > VPII;
108 typedef map< int, int > MII;
109 typedef map< int, string > MIS;
110 typedef map< int, PII > MIPII;
111 typedef map< PII, int > MPIII;
112 typedef map< string, int > MSI;
113 typedef map< string, string > MSS;
114 typedef map< PII, string > MPIIS;
115 typedef map< PII, PII > MPIIPII;
116 typedef multimap< int, int > MMII;
117 typedef multimap< string, int > MMSI;
118 //typedef unordered_map< int, int > uMII;
119 typedef pair< LL, LL > PLL;
120 typedef vector< LL > VL;
121 typedef vector< VL > VVL;
122 typedef priority_queue< int > PQIMax;
123 typedef priority_queue< int, VI, greater< int > > PQIMin;
124 const double EPS = 1e-8;
125 const LL inf = 0x7fffffff;
126 const LL infLL = 0x7fffffffffffffffLL;
127 const LL mod = 1e9 + 7;
128 const int maxN = 1e5 + 7;
129 const LL ONE = 1;
130 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
131 const LL oddBits = 0x5555555555555555;
132
133 int N;
134 int P[maxN]; // P数组为1~N的一组排列
135 int pos[maxN]; // pos[i]为i在P数组中的位置
136 int leftFirstMaxPos[maxN]; // leftFirstMaxPos[i]为i位置的数左边第一个比i大的数的位置,没有则为0
137 int rightFirstMaxPos[maxN]; // rightFirstMaxPos[i]为i位置的数右边第一个比i大的数的位置,没有则为N+1
138 LL ans = 0;
139
140 int main(){
141 //freopen("MyOutput.txt","w",stdout);
142 //freopen("input.txt","r",stdin);
143 //INIT();
144 scanf("%d", &N);
145 For(i, 1, N) {
146 scanf("%d", &P[i]);
147 pos[P[i]] = i;
148 // 后面是从1开始枚举,而1是最小的,所以1的左右最近大正好是相邻的两个数
149 leftFirstMaxPos[i] = i - 1;
150 rightFirstMaxPos[i] = i + 1;
151 }
152
153 // 计算每个数i为第二大的时候对答案的贡献
154 For(i, 1, N) {
155 int Lpos1 = leftFirstMaxPos[pos[i]]; // i左边第一个比他大的数的位置
156 int Lpos2 = leftFirstMaxPos[Lpos1]; // i左边第二个比他大的数的位置
157 int Rpos1 = rightFirstMaxPos[pos[i]]; // i右边第一个比他大的数的位置
158 int Rpos2 = rightFirstMaxPos[Rpos1]; // i右边第二个比他大的数的位置
159
160 if(Lpos1 != 0) ans += ONE * i * (Lpos1 - Lpos2) * (Rpos1 - pos[i]);
161 if(Rpos1 != N + 1) ans += ONE * i * (pos[i] - Lpos1) * (Rpos2 - Rpos1);
162
163 // 更新
164 /*
165 例如这个片段:
166 1 2 3 4 5 6 7
167 P 6 3 1 2 5 4 7
168 此时以i = 1为例
169 本来leftFirstMaxPos[4] = 3, rightFirstMaxPos[2] = 3的
170 计算完i = 1的贡献之后, leftFirstMaxPos[4] = 2, rightFirstMaxPos[2] = 4
171 这一步更新相当于把1给砍掉了,剩下的数中2又是最小的数,于是2的左右最近大又正好是相邻的两个数
172 没错,非常像链表删除一个值的操作
173 leftFirstMaxPos就相当于静态链表的前驱数组
174 rightFirstMaxPos就相当于静态链表的后继数组
175 */
176 leftFirstMaxPos[Rpos1] = Lpos1;
177 rightFirstMaxPos[Lpos1] = Rpos1;
178 }
179
180 printf("%lld", ans);
181 return 0;
182 }
转载于:https://www.cnblogs.com/zaq19970105/p/11486861.html