树状数组求逆序对
我果然还是太NAIVE了
一直没有学到这么搞基的求逆序对数方法
其实就是有点像计数排序
先开一个巨大的树状数组
然后插入一个数的时候看一下SUM
用当前位置减一下SUM就可以了
//其实就是求和
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cmath> 7 using namespace std; 8 long long read() { 9 long long x = 0, f = 1; char ch = getchar(); 10 while (!isdigit(ch)) { 11 if (ch == '-') f = -1; ch = getchar(); 12 } 13 while (isdigit(ch)) { 14 x = x * 10 + ch - '0'; ch = getchar(); 15 } 16 return x * f; 17 } 18 19 /*------------------------------template------------------------------*/ 20 21 #define nxt(x) (x & -x) 22 #define M 500050 23 typedef pair<int, int> pr; 24 int a[M << 1]; 25 long long ans = 0; 26 int c[M], b[M]; 27 int main () { 28 int n = read(), x; 29 memset(c, 0, sizeof(c)); 30 memset(a, 0, sizeof(a)); 31 for (int i = 1; i <= n; i ++) { 32 b[i] = c[i] = read(); 33 } 34 sort(b + 1, b + 1 + n); 35 int tot = unique(b + 1, b + 1 + n) - b; 36 for (int i = 1; i <= n; i ++) { 37 c[i] = lower_bound(b + 1, b + tot, c[i]) - b; 38 } 39 for (int i = 1; i <= n; ans += i ++) { 40 x = c[i]; 41 for (int j = x; j <= n; j += nxt(j)) { 42 a[j] ++; 43 } 44 for (int j = x; j > 0; j -= nxt(j)) { 45 ans -= a[j]; 46 } 47 //printf("%d %d %d\n", x, i, ans); 48 } 49 printf("%lld\n", ans); //打成%d调了半天23333333 50 return 0; 51 }
于是就可以去做http://www.lydsy.com/JudgeOnline/problem.php?id=2141了哈哈哈
终于码完了23333
由于代码能力还不够最后还是对着调的。。。
题解：http://blog.csdn.net/PoPoQQQ/article/details/40373903
1 #include <bits/stdc++.h> 2 using namespace std; 3 long long read() { 4 long long x = 0, f = 1; char ch = getchar(); 5 while (!isdigit(ch)) { 6 if (ch == '-') f = -1; ch = getchar(); 7 } 8 while (isdigit(ch)) { 9 x = x * 10 + ch - '0'; ch = getchar(); 10 } 11 return x * f; 12 } 13 14 /*------------------------------template------------------------------*/ 15 16 #define M 20200 17 #define nxt(x) (x & -x) 18 using namespace std; 19 20 int n, m, ans, tot, block; 21 int a[M]; 22 pair<int, int> b[M]; 23 int pre[M], cnt[200][M]; 24 25 void up(int c[], int x) { 26 for(; x <= n; x += nxt(x)) ++c[x]; 27 } 28 29 void down(int c[], int x) { 30 for(; x <= n; x += nxt(x)) --c[x]; 31 } 32 33 int calc(int c[], int x) { 34 int re = 0; 35 for(; x; x -= nxt(x)) re+=c[x]; 36 return re; 37 } 38 39 int main() { 40 int x, y; 41 n = read(); 42 for(int i = 1; i <= n; i ++) b[i] = make_pair(read(), i); 43 sort(b + 1, b + n + 1); 44 for(int i = 1; i <= n; i ++) { 45 if (b[i].first != b[i - 1].first) ++tot; 46 a[b[i].second] = tot; 47 } 48 for(int i = n; i; i --) ans += calc(pre, a[i] - 1), up(pre, a[i]); 49 printf("%d\n", ans); 50 //Get the first ans 51 block = static_cast<int>(sqrt(n) + 1e-7); 52 for(int i = 1; i <= n; i ++) up(cnt[(i-1) / block], a[i]); 53 //Blocking init 54 m = read(); 55 for(int i = 1; i <= m; i ++) { 56 x = read(); y = read(); 57 if (x > y) swap(x, y); 58 //查询(x, y)区间内有多少个数比a[x]和a[y]小 59 int b1 = (x - 1) / block + 1; 60 int b2 = (y - 1) / block - 1; 61 if (b1 <= b2) { 62 for(int j = b1; j <= b2; j ++) { 63 ans -= calc(cnt[j], a[x] - 1); 64 ans += calc(cnt[j], n) - calc(cnt[j], a[x]); 65 ans += calc(cnt[j], a[y] - 1); 66 ans -= calc(cnt[j], n) - calc(cnt[j], a[y]); 67 } 68 for(int j = x + 1; j <= b1 * block; j ++) { 69 if (a[j] < a[x]) --ans; 70 if (a[j] > a[x]) ++ans; 71 if (a[j] < a[y]) ++ans; 72 if (a[j] > a[y]) --ans; 73 } 74 for(int j = (b2 + 1) * block + 1; j < y; j ++) { 75 if (a[j]<a[x]) --ans; 76 if (a[j]>a[x]) ++ans; 77 if (a[j]<a[y]) ++ans; 78 if (a[j]>a[y]) --ans; 79 } 80 } 81 else { 82 //暴力 83 for(int j = x + 1; j <= y - 1; j ++) { 84 if (a[j] < a[x]) --ans; 85 if (a[j] > a[x]) ++ans; 86 if (a[j] < a[y]) ++ans; 87 if (a[j] > a[y]) --ans; 88 } 89 } 90 if (a[x] < a[y]) ++ans; 91 else if (a[x] > a[y]) --ans; 92 printf("%d\n", ans); 93 down(cnt[(x-1) / block], a[x]); 94 down(cnt[(y-1) / block], a[y]); 95 swap(a[x], a[y]); 96 up(cnt[(x-1) / block], a[x]); 97 up(cnt[(y-1) / block], a[y]); 98 } 99 }
