「JSOI2011」任务调度

「JSOI2011」任务调度

传送门
一开始还在想写平衡树，看到 ( ext{TRANS}) 操作后就晓得要用可并堆了。
这题好像就是个可并堆的板子题？？？

• ADD 直接往对应的对里面加元素
• DEC 在对应的堆里面找到这个元素，讨论一下它是不是根节点，然后抠出来重新加进去
• TRANS 合并两个堆
• MIN 查堆顶的值
• WORK 讨论一下根节点和它儿子的大小关系来判 ERROR 的情况，然后就和 DEC 一样了

参考代码：

#include <algorithm>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 502, __ = 300005;

int n, m, q, rt[_];
int fa[__], val[__], dis[__], ch[2][__];

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

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

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(m), read(q);
    char s[10];
    for (rg int x, y, z, i = 1; i <= q; ++i) {
    	scanf("%s", s);
    	if (s[0] == 'A') read(x), read(y), read(z), val[y] = z, rt[x] = merge(rt[x], y);
    	if (s[0] == 'D') read(x), read(y), read(z), update(x, y, -z);
    	if (s[0] == 'T') read(x), read(y), rt[y] = merge(rt[x], rt[y]), rt[x] = 0;
    	if (s[0] == 'M') read(x), printf("%d
", val[rt[x]]);
    	if (s[0] == 'W') {
    	    read(x), read(y);
    	    if (ch[0][rt[x]] && val[ch[0][rt[x]]] == val[rt[x]]) { puts("ERROR"); continue ; }
    	    if (ch[1][rt[x]] && val[ch[1][rt[x]]] == val[rt[x]]) { puts("ERROR"); continue ; }
    	    update(x, rt[x], y);
    	}
    }
    return 0;
}