CSU 1555 Inversion Sequence 给出逆序数求排列 splay

题目链接：点击打开链接

题意：

给出逆序数的值，求原序列（一个1-N的排列）

1, 2, 0, 1, 0 表示1的逆序数是1，2的逆序数是2，3的逆序数是0···

思路：

从最后一个数开始插，每次插到当前序列的第a[i]个数。。

splay模拟

== 这个方法比较直(wu)观(nao)，别的方法并没有想出来。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if(c=getchar(),c==EOF) return 0;
	while(c!='-'&&(c<'0'||c>'9')) c=getchar();
	sgn=(c=='-')?-1:1;
	ret=(c=='-')?0:(c-'0');
	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
	ret*=sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) {
        putchar('-');
        x = -x;
    }
    if(x>9) pt(x/10);
    putchar(x%10+'0');
}
using namespace std;
inline int Mid(int a,int b){return (a+b)>>1;}
#define N 100010
#define L(x) tree[x].ch[0]
#define R(x) tree[x].ch[1]
#define Siz(x) tree[x].siz
#define Father(x) tree[x].fa
#define Max(x) tree[x].max
#define Val(x) tree[x].val
#define Pt(x) tree[x].pt()
struct node{
    int ch[2], siz, fa;
    int max, val;
    void pt(){printf("val:%d max:%d siz:%d fa:%d child{%d,%d}\n", val,max,siz,fa,ch[0],ch[1]);}
}tree[N*2];
int tot, root;
void Newnode(int &id, int val, int fa, int siz = 1){
    id = ++tot;
    L(id) = R(id) = 0;
    Father(id) = fa;
    Siz(id) = siz;
    Max(id) = Val(id) = val;
}

void push_up(int id){
    Siz(id) = Siz(L(id)) + Siz(R(id)) +1;
    Max(id) = max(Max(R(id)), Max(L(id)));
    Max(id) = max(Val(id), Max(id));
}
void push_down(int id){}

void Rotate(int id, int kind){
    int y = Father(id);
    push_down(y); push_down(id); //here
    tree[y].ch[kind^1] = tree[id].ch[kind];
    Father(tree[id].ch[kind]) = y;
    if(Father(y))
        tree[Father(y)].ch[R(Father(y))==y] = id;
    Father(id) = Father(y);
    Father(y) = id;
    tree[id].ch[kind] = y;
    push_up(y);
}
void splay(int id, int goal){
    push_down(id);
    while(Father(id) != goal){
        int y = Father(id);
        if(Father(y) == goal)
            Rotate(id, L(y)==id);
        else
        {
            int kind = L(Father(y)) == y;
            if(tree[y].ch[kind] == id)
            {
                Rotate(id, kind^1);
                Rotate(id, kind);
            }
            else
            {
                Rotate(y, kind);
                Rotate(id,kind);
            }
        }
    }
    push_up(id);
    if(goal == 0)root = id;
}
int Get_kth(int kth, int sor){//找到在sor后面的第k个数
    push_down(sor);
    int id = sor;
    while(Siz(L(id)) != kth){
        if(Siz(L(id)) > kth)
            id = L(id);
        else
        {
            kth -= (Siz(L(id))+1);
            id = R(id);
        }
        push_down(id);
    }
    return id;
}
void init(){
	Father(0) = L(0) = R(0) = Siz(0) = 0;
	Max(0) = 0;
	tot = 0;
	Newnode(root, 0, 0);
	Newnode(R(root), 0, root);
	push_up(root);
}
void debug(int x){
	printf("%d:\n", x);
	Pt(x);
	if(L(x)){
		printf("L:");
		debug(L(x));
		printf("return to %d:\n", x);
	}
	if(R(x)){
		printf("R:");
		debug(R(x));
		printf("return to %d:\n", x);
	}
}
void insert(int pos, int val){
	splay(1, 0);
	int u = Get_kth(pos, 1);
//	if(pos == 2){cout<<"=="; debug(root);}
	int v = Get_kth(pos+1, 1);
//	printf("在(%d,%d)之间:", u, v);	Pt(u); Pt(v);	puts("*****");
	splay(u, 0);
	splay(v, root);
//	if(pos == 2){cout<<"=="; debug(root);}
	Newnode(L(v), val, v);
	push_up(v);
	push_up(u);
//	printf("[%d,%d]\n", u, v);
}

int n;
int a[100005];
vector<int>G;
void dfs(int u){
	if(L(u))
		dfs(L(u));
	G.push_back(Val(u));
	if(R(u))
		dfs(R(u));
}
int main() {
    while(cin>>n){
    	init();
	//		puts("init debug begin:");debug(root);
		for(int i = 1; i <= n; i++)rd(a[i]);
		bool ok = true;
		for(int i = n; i; i--){
			if(a[i]>n-i){ok = false; break;	}
			insert(a[i] , i);
	//		printf("   %d debug begin:", i);debug(root);
		}
		if(false == ok){ puts("No solution");continue; }
		G.clear();
		dfs(root);
		for(int i = 1; i < G.size() -1; i++)
		printf("%d%c", G[i], i+2==(int)G.size()?'\n':' ');
    }
    return 0;
}
/*
1
7
0 1 1 1 0 4 1

*/

CSU 1555 Inversion Sequence 给出逆序数求排列 splay

原文：http://blog.csdn.net/qq574857122/article/details/44892147