P1908 逆序对
1.大雪球2.E. Final Countdown3.D. Divisible Pairs4.F. Chat Screenshots5.G. One-Dimensional Puzzle6.P8775 [蓝桥杯 2022 省 A] 青蛙过河7.Watering an Array8.Dora and Search9.P1137 旅行计划10.P3067 [USACO12OPEN] Balanced Cow Subsets G11.P4799 [CEOI2015 Day2] 世界冰球锦标赛12.A. Brick Wall13.B. Minimize Inversions14.C. XOR-distance15.A. Moving Chips16.B. Monsters Attack!17.C. Find B18.D. Slimes19.C. Turtle Fingers: Count the Values of k20.D. Turtle Tenacity: Continual Mods21.D. Vlad and Division22.C. LR-remainders23.P5322 [BJOI2019] 排兵布阵24.P2946 [USACO09MAR] Cow Frisbee Team S25.P1156 垃圾陷阱26.P1064 [NOIP2006 提高组] 金明的预算方案27.P3047 [USACO12FEB] Nearby Cows G28.P5020 [NOIP2018 提高组] 货币系统29.P3957 [NOIP2017 普及组] 跳房子30.P1550 [USACO08OCT] Watering Hole G31.P1525 [NOIP2010 提高组] 关押罪犯32.P1379 八数码难题33.P6121 [USACO16OPEN] Closing the Farm G34.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G35.P9632 [ICPC2020 Nanjing R] K Co-prime Permutation36.P9825 [ICPC2020 Shanghai R] Fibonacci37.P2034 选择数字38.P7072 [CSP-J2020] 直播获奖39.P1714 切蛋糕40.P4147 玉蟾宫41.P2866 [USACO06NOV] Bad Hair Day S42.P2032 扫描43.P1884 [USACO12FEB] Overplanting S44.D. Lonely Mountain Dungeons45.F. Microcycle46.E. Increasing Subsequences__247.D. Valiant's New Map48.Array Repetition49.F. Rudolf and Imbalance50.D. Exam in MAC51.G. Rudolf and Subway52.C. Ehab and Path-etic MEXs53.A. String Transformation 154.D. Solve The Maze55.B. Array Fix56.B. Equal XOR57.F. Maximum White Subtree58.P1652 圆59.P2181 对角线60.P1325 雷达安装61.P3194 [HNOI2008] 水平可见直线62.B - Make Many Triangles63.[ARC172A] Chocolate64.C. Theofanis' Nightmare65.D. Birthday Gift66.C. Arrow Path67.P1265 公路修建68.P1284 三角形牧场69.Kirill and Mushrooms70.C. Sasha and the Casino71.P1470 [USACO2.3] 最长前缀 Longest Prefix72.P5536 【XR-3】核心城市73.P5836 [USACO19DEC] Milk Visits S74.P4391 [BOI2009] Radio Transmission 无线传输75.P3435 [POI2006] OKR-Periods of Words76.P5960 【模板】差分约束77.P3052 [USACO12MAR] Cows in a Skyscraper G78.E. Long Inversions79.C. Deep Down Below80.C. Rings81.F. Unfair Game82.G. GCD on a grid83.C. Inversion Graph84.Increase Subarray Sums85.D. Cross Coloring86.M. Triangle Construction87.D2. Set To Max (Hard Version)88.C. Array Game89.F. Alex's whims90.Queue Sort
91.P1908 逆序对
92.P5094 [USACO04OPEN] MooFest G 加强版93.P5149 会议座位94.P10337 [UESTCPC 2024] 操作序列95.P10330 [UESTCPC 2024] 黑白珠串96.P10252 线性变换97.P3901 数列找不同98.P10288 [GESP样题 八级] 区间99.P5677 [GZOI2017] 配对统计题解1
本质:求一个数前面有几个数大于它,我们把序列分成几段,然后对每段分别进行排序,然后找出这个数在前面已经排好序中的序列里有几个大于它
code
#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[500005], b[500005], c[500005]; // a-original, b-replace, c-new
ll ans = 0;
inline void read(ll &x) {
x = 0;
ll flag = 1;
char c = getchar();
while(c < '0' || c > '9'){
if(c == '-')flag = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = (x << 3) + (x << 1) + (c ^ 48);
c = getchar();
}
x *= flag;
}
inline void write(ll x)
{
if(x < 0){
putchar('-');
x = -x;
}
if(x > 9)
write(x / 10);
putchar(x % 10 + '0');
}
void bg(ll l, ll r)
{
if (l == r) return;
ll mid = (l + r) / 2;
bg(l, mid);
bg(mid + 1, r);
ll p1 = l, p2 = mid + 1, p3 = l;
while (p1 <= mid && p2 <= r)
{
if (c[p1] <= c[p2]) b[p3++] = c[p1++];//为什么要等于?因为等于不算逆序对
else
{
b[p3++] = c[p2++];
ans += mid - p1 + 1;
}
}
while (p1 <= mid) b[p3++] = c[p1++];
while (p2 <= r) b[p3++] = c[p2++];
for (ll i = l; i <= r; i++) c[i] = b[i];
return;
}
int main()
{
ll n;
read(n);
for (ll i = 1; i <= n; i++)
{
read(a[i]);
c[i]=a[i];
}
bg(1, n);
write(ans);
return 0;
}
题解2:
设遍历到
code
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#define lowbit(x) ((x)&(-x))
using namespace __gnu_pbds;
using namespace std;
int a[500005],c[500005];
int b[500005]={0};
int n;
int len=0;
int query(int x)
{
int sum=0;
while(x)
{
sum+=b[x];
x-=lowbit(x);
}
return sum;
}
void update(int x)
{
while(x<=len)
{
b[x]++;
x+=lowbit(x);
}
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
c[i]=a[i];
}
sort(a+1,a+n+1);
gp_hash_table<int, int> id;
for(int i=1;i<=n;i++)
{
if(!id[a[i]]) id[a[i]]=++len;
}
int ans=0;
for(int i=1;i<=n;i++)
{
update(id[c[i]]);
ans+=i-query(id[c[i]]);//i represent the vals that have appeared,query() function gets the number of vals smaller than a[i] that have appeared
}
cout<<ans;
return 0;
}
合集:
思维题
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