[题解] [JSOI2011] 任务调度

题解

题面

左偏树练习题吧
改权值的操作就是把这个点扯出来, 左右儿子合并后接到这个点的父亲上去, 然后再把这个点重新塞进左偏树里就行了
至于这部操作为什么不要更新 dis , 可能是因为 dis 最多只会变 1 , 左偏的性质还是存在吧

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
const int N = 300005; 
using namespace std;

int n, m, T, rt[505];
char s[105]; 
struct node
{
    int val, dis, fa, ch[2]; 
} t[N]; 

template < typename T >
inline T read()
{
    T x = 0, w = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * w; 
}

int merge(int x, int y)
{
    if(!x || !y) return x + y;
    if(t[x].val > t[y].val) swap(x, y);
    t[x].ch[1] = merge(t[x].ch[1], y);
    if(t[t[x].ch[1]].dis > t[t[x].ch[0]].dis) swap(t[x].ch[0], t[x].ch[1]);
    t[x].dis = t[t[x].ch[1]].dis + 1;
    t[t[x].ch[0]].fa = t[t[x].ch[1]].fa = x;
    return x; 
}

void modify(int x, int y, int z)
{
    int f = t[y].fa, tmp = merge(t[y].ch[0], t[y].ch[1]);
    if(rt[x] == y) rt[x] = tmp; 
    else t[f].ch[t[f].ch[1] == y] = tmp;
    t[tmp].fa = f; t[y].val = z, t[y].fa = 0;
    t[y].ch[0] = t[y].ch[1] = 0;
    rt[x] = merge(rt[x], y); 
}

int main()
{
    n = read <int> (), m = read <int> (), T = read <int> ();
    int x, y, z; 
    while(T--)
    {
	scanf("%s", s + 1); 
	if(s[1] == 'A')
	{
	    x = read <int> (), y = read <int> (), z = read <int> ();
	    modify(x, y, z);
//	    t[y].val = z, rt[x] = merge(rt[x], y); 
	}
	else if(s[1] == 'D')
	{
	    x = read <int> (), y = read <int> (), z = read <int> ();
	    modify(x, y, t[y].val - z); 
	}
	else if(s[1] == 'T')
	{
	    x = read <int> (), y = read <int> ();
	    rt[y] = merge(rt[y], rt[x]), rt[x] = 0; 
	}
	else if(s[1] == 'M')
	    x = read <int> (), printf("%d
", t[rt[x]].val); 
	else if(s[1] == 'W')
	{
	    x = read <int> (), y = read <int> ();
	    if((t[rt[x]].ch[0] && t[rt[x]].val == t[t[rt[x]].ch[0]].val) || (t[rt[x]].ch[1] && t[rt[x]].val == t[t[rt[x]].ch[1]].val))
		puts("ERROR"); 
	    else modify(x, rt[x], t[rt[x]].val + y); 
	}
    }
    return 0; 
}