Runing on judge 1 5 9 28……Accepted
处理字符串，根据字符串的字典序对其进行赋予权值，接近a的权值大些，然后在按转置的字符串X' 对 X（原串）进行排序，再然后就是很裸的树状数组模板求逆序数了。
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Little Lovro likes to play games with words. During the last few weeks he realized that some words don't like each other.
The words A and B don't like each other if the word A is lexicographically before the word B, but the word B' is lexicographically before the word A', where X' stands for the word X reversed (if X="kamen" then X'="nemak"). For example, the words "lova" and "novac" like each other, but the words "aron" and "sunce" don't like each other.
Given some set of the words, we define the degree of chaos of the set as the number of pairs of different words that don't like each other.
Write a program that, given a set of words, finds the chaos degree for the set.
Input
The first line of input contains an integer N, 2 ≤ N ≤ 100 000.
Each of the following N lines contains one word – a sequence of at most 10 lowercase letters of the English alphabet ('a'-'z'). There will be no two identical words in the set.
Output
The first and only line of output should contain a single integer – the chaos degree of the given set of words.
Note: use 64-bit signed integer type (int64 in Pascal, long long in C/C++).
Sample
input
2
lopta
kugla
output
0
input
4
lova
novac
aron
sunce
output
3
input
14
branimir
vladimir
tom
kruz
bred
pit
zemlja
nije
ravna
ploca
ko
je
zapalio
zito
output
48
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#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#define N 100007
using namespace std;
long long bit[N];
int n;
struct Str {
char str[12];
int p; //权重
bool operator < (const Str rhs) const { //两字符串的逆串比较大小, z->a
int len1 = strlen(str), len2 = strlen(rhs.str);
while (len1 >= 0 && len2 >= 0) {
if (str[len1] == rhs.str[len2])
len1--, len2--;
else
return (str[len1] > rhs.str[len2]);
}
return len1 >= len2;
}
}a[N];
bool cmp(Str lhs, Str rhs) {
return strcmp(lhs.str, rhs.str) > 0; //从大到小排序z->a
}
inline int lowbit(int x) {
return x & (-x);
}
void add(int pos, int val) {
while (pos <= n) {
bit[pos] += val;
pos += lowbit(pos);
}
}
long long sum(int pos) {
long long res = 0;
while (pos > 0) {
res += bit[pos];
pos -= lowbit(pos);
}
return res;
}
int main()
{
while (~scanf("%d", &n)) {
memset(bit, 0, sizeof bit);
memset(a, 0, sizeof a);
for (int i = 1; i <= n; ++i) {
scanf("%s", a[i].str);
}
sort(a + 1, a + n + 1, cmp);
for (int i = 1; i <= n; ++i)
a[i].p = i;
sort(a + 1, a + n + 1);
long long ans = 0;
for (int i = 1; i <= n; ++i) {
ans += (i - sum(a[i].p) - 1);
add(a[i].p, 1);
}
printf("%lld\n", ans);
}
return 0;
}
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