[SCOI2011]棘手的操作 题解
题目地址:洛谷:【P3273】[SCOI2011]棘手的操作 – 洛谷、BZOJ:Problem 2333. — [SCOI2011]棘手的操作
题目描述
有N个节点,标号从1到N,这N个节点一开始相互不连通。第i个节点的初始权值为a[i],接下来有如下一些操作:
U x y
: 加一条边,连接第x个节点和第y个节点A1 x v
: 将第x个节点的权值增加vA2 x v
: 将第x个节点所在的连通块的所有节点的权值都增加vA3 v
: 将所有节点的权值都增加vF1 x
: 输出第x个节点当前的权值F2 x
: 输出第x个节点所在的连通块中,权值最大的节点的权值F3
: 输出所有节点中,权值最大的节点的权值
输入输出格式
输入格式:
输入的第一行是一个整数N,代表节点个数。接下来一行输入N个整数,a[1], a[2], …, a[N],代表N个节点的初始权值。再下一行输入一个整数Q,代表接下来的操作数。最后输入Q行,每行的格式如题目描述所示。
输出格式:
对于操作F1, F2, F3,输出对应的结果,每个结果占一行。
输入输出样例
输入样例#1:
3 0 0 0 8 A1 3 -20 A1 2 20 U 1 3 A2 1 10 F1 3 F2 3 A3 -10 F3
输出样例#1:
-10 10 10
说明
对于30%的数据,保证 N<=100,Q<=10000
对于80%的数据,保证 N<=100000,Q<=100000
对于100%的数据,保证 N<=300000,Q<=300000
对于所有的数据,保证输入合法,并且 -1000<=v, a[1], a[2], …, a[N]<=1000
题解
解法1:线段树 + 并查集
考虑离线解决这个题,我们可以把操作中同一连通块的元素连续地排列在一个序列中,这样连通块操作就变成了区间操作,这个可以用线段树维护区间最大值。
至于如何连续排列,可以考虑用链表来存储当前连通块序列,然后合并连通块的时候合并链表,这样就自然成为连续的一段了。最后再遍历链表给节点加编号就好。
解法2:可并堆
使用可并堆做这个题目就很裸了,也不需要离线。这里我写的是左偏树。
分开管理两类可并堆,一类是联通块的大根堆,一类是所有联通块堆顶元素的大根堆(换句话说,这是一个维护全局的堆)。
U
操作:将两个节点所在的堆合并,删去全局堆中消失的那个堆的对顶。
A1
操作:将x所在堆中x元素删去,更新x元素的值,再插回去。
A2
操作:将x所在堆全局增量加上v。堆全局增量可以在堆顶维护,需要统计这个结点的真实值的时候从该点一路上跳,跳到一个父亲就将该父亲的增量加入真实值。(因为堆并过程中增量仍然在被合并的堆顶)
A3
操作:维护总全局增量加v。
F1
操作:查询x的真实值。
F2
操作:查询x所在堆堆顶的真实值。
F3
操作:查询全局堆堆顶的真实值。
这个解法的代码我借鉴了远航之曲的实现思路。
代码
解法1
// Code by KSkun, 2018/2
#include <cstdio>
#include <algorithm>
struct io {
char buf[1 << 26], *op;
io() {
fread(op = buf, 1, 1 << 26, stdin);
}
inline int readint() {
register int res = 0, neg = 1;
while(*op < '0' || *op > '9') if(*op++ == '-') neg = -1;
while(*op >= '0' && *op <= '9') res = res * 10 + *op++ - '0';
return res * neg;
}
inline char readchar() {
return *op++;
}
} ip;
#define readint ip.readint
#define readchar ip.readchar
inline bool isop(char c) {
return c == 'A' || c == 'U' || c == 'F';
}
inline char readop() {
char c;
while(!isop(c = readchar()));
return c;
}
int n, a[300005], q, opop1[300005], opx[300005], opy[300005], idx[300005], inv[300005], tot = 1;
char opop[300005];
// Linked List
int pre[300005], nxt[300005], head[300005], tail[300005];
// Union-Find
int fa[300005];
inline int find(int x) {
int r = x;
while(fa[r] != r) r = fa[r];
while(fa[x] != x) {
int ofa = fa[x];
fa[x] = r;
x = ofa;
}
return r;
}
// Seg Tree
#define lch o << 1
#define rch o << 1 | 1
#define mid ((l + r) >> 1)
int tree[1200005], lazy[1200005];
inline void build(int o, int l, int r) {
if(l == r) {
tree[o] = a[inv[l]];
return;
}
build(lch, l, mid);
build(rch, mid + 1, r);
tree[o] = std::max(tree[lch], tree[rch]);
}
inline void pushdown(int o) {
if(lazy[o] != 0) {
lazy[lch] += lazy[o];
lazy[rch] += lazy[o];
tree[lch] += lazy[o];
tree[rch] += lazy[o];
lazy[o] = 0;
}
}
inline void add(int o, int l, int r, int ll, int rr, int v) {
if(l == ll && r == rr) {
tree[o] += v;
lazy[o] += v;
return;
}
pushdown(o);
if(rr <= mid) {
add(lch, l, mid, ll, rr, v);
} else if(ll > mid) {
add(rch, mid + 1, r, ll, rr, v);
} else {
add(lch, l, mid, ll, mid, v);
add(rch, mid + 1, r, mid + 1, rr, v);
}
tree[o] = std::max(tree[lch], tree[rch]);
}
inline int query(int o, int l, int r, int ll, int rr) {
if(l == ll && r == rr) {
return tree[o];
}
pushdown(o);
if(rr <= mid) {
return query(lch, l, mid, ll, rr);
} else if(ll > mid) {
return query(rch, mid + 1, r, ll, rr);
} else {
return std::max(query(lch, l, mid, ll, mid), query(rch, mid + 1, r, mid + 1, rr));
}
}
int main() {
n = readint();
for(int i = 1; i <= n; i++) {
a[i] = readint();
fa[i] = head[i] = tail[i] = i;
}
q = readint();
for(int i = 1; i <= q; i++) {
opop[i] = readop();
if(opop[i] == 'U') {
opx[i] = readint();
opy[i] = readint();
int f1 = find(opx[i]), f2 = find(opy[i]);
if(f1 != f2) {
fa[f2] = f1;
pre[head[f2]] = tail[f1];
nxt[tail[f1]] = head[f2];
tail[f1] = tail[f2];
}
} else if(opop[i] == 'A') {
opop1[i] = readint();
if(opop1[i] < 3) {
opx[i] = readint();
}
opy[i] = readint();
} else if(opop[i] == 'F') {
opop1[i] = readint();
if(opop1[i] < 3) {
opx[i] = readint();
}
}
}
for(int i = 1; i <= n; i++) {
if(!pre[i]) {
for(int j = i; j; j = nxt[j]) {
idx[j] = tot++;
inv[idx[j]] = j;
}
}
}
build(1, 1, n);
for(int i = 1; i <= n; i++) {
fa[i] = head[i] = tail[i] = i;
}
for(int i = 1; i <= q; i++) {
if(opop[i] == 'U') {
int f1 = find(idx[opx[i]]), f2 = find(idx[opy[i]]);
if(f1 != f2) {
fa[f2] = f1;
if(head[f1] < head[f2]) {
tail[f1] = tail[f2];
} else {
head[f1] = head[f2];
}
}
} else if(opop[i] == 'A') {
if(opop1[i] == 1) {
add(1, 1, n, idx[opx[i]], idx[opx[i]], opy[i]);
} else if(opop1[i] == 2) {
int f = find(idx[opx[i]]);
add(1, 1, n, head[f], tail[f], opy[i]);
} else {
add(1, 1, n, 1, n, opy[i]);
}
} else if(opop[i] == 'F') {
if(opop1[i] == 1) {
printf("%d\n", query(1, 1, n, idx[opx[i]], idx[opx[i]]));
} else if(opop1[i] == 2) {
int f = find(idx[opx[i]]);
printf("%d\n", query(1, 1, n, head[f], tail[f]));
} else {
printf("%d\n", query(1, 1, n, 1, n));
}
}
}
return 0;
}
解法2
// Code by KSkun, 2018/2
#include <cstdio>
#include <queue>
struct io {
char buf[1 << 26], *op;
io() {
fread(op = buf, 1, 1 << 26, stdin);
}
inline int readint() {
register int res = 0, neg = 1;
while(*op < '0' || *op > '9') if(*op++ == '-') neg = -1;
while(*op >= '0' && *op <= '9') res = res * 10 + *op++ - '0';
return res * neg;
}
inline char readchar() {
return *op++;
}
} ip;
#define readint ip.readint
#define readchar ip.readchar
inline bool isop(char c) {
return c == 'U' || c == 'A' || c == 'F';
}
inline void readop(char* str) {
char c;
while(!isop(c = readchar()));
str[0] = c;
if(c == 'A' || c == 'F') str[1] = readchar();
}
int n, m, x, y, z, add = 0;
char op[5];
struct LeftistTree {
int dis[300005], fa[300005], ch[300005][2], val[300005], add[300005], root;
inline int toadd(int x) {
int res = 0;
while(x = fa[x]) res += add[x];
return res;
}
inline void clear(int x) {
fa[x] = ch[x][0] = ch[x][1] = 0;
}
inline void pushdown(int x) {
if(ch[x][0]) {
val[ch[x][0]] += add[x];
add[ch[x][0]] += add[x];
}
if(ch[x][1]) {
val[ch[x][1]] += add[x];
add[ch[x][1]] += add[x];
}
add[x] = 0;
}
inline int merge(int x, int y) {
if(!x) return y;
if(!y) return x;
if(val[x] < val[y]) std::swap(x, y);
pushdown(x);
ch[x][1] = merge(ch[x][1], y);
fa[ch[x][1]] = x;
if(dis[ch[x][0]] < dis[ch[x][1]]) std::swap(ch[x][0], ch[x][1]);
dis[x] = dis[ch[x][1]] + 1;
return x;
}
inline int froot(int x) {
while(fa[x]) x = fa[x];
return x;
}
inline void delet(int x) {
pushdown(x);
int f = merge(ch[x][0], ch[x][1]);
fa[f] = fa[x];
if(fa[x]) ch[fa[x]][ch[fa[x]][0] == x ? 0 : 1] = f;
int t = fa[x];
if(!t) {
root = f;
return;
}
while(t) {
if(dis[ch[t][0]] < dis[ch[t][1]]) std::swap(ch[t][0], ch[t][1]);
if(dis[t] == dis[ch[t][1]] + 1) return;
dis[t] = dis[ch[t][1]] + 1;
if(!fa[t]) root = t;
t = fa[t];
}
}
inline void addtree(int x, int v) {
int rt = froot(x);
val[rt] += v;
add[rt] += v;
}
inline int addpoint(int x, int v) {
int rt = froot(x);
if(rt == x) {
if(!ch[x][0] && !ch[x][1]) {
val[x] += v;
return x;
} else {
rt = ch[x][0] ? ch[x][0] : ch[x][1];
}
}
delet(x);
val[x] += v + toadd(x);
clear(x);
return merge(froot(rt), x);
}
inline void build() {
std::queue<int> que;
for(int i = 1; i <= n; i++) {
que.push(i);
}
while(que.size() > 1) {
int x = que.front();
que.pop();
int y = que.front();
que.pop();
que.push(merge(x, y));
}
root = que.front();
}
};
LeftistTree lt1, lt2;
int main() {
n = readint();
for(int i = 1; i <= n; i++) {
lt1.val[i] = lt2.val[i] = readint();
}
lt1.dis[0] = lt2.dis[0] = -1;
lt2.build();
m = readint();
for(int i = 1; i <= m; i++) {
readop(op);
if(op[0] == 'U') {
x = readint();
y = readint();
int fx = lt1.froot(x), fy = lt1.froot(y), rt;
if(fx != fy) {
rt = lt1.merge(fx, fy);
if(rt == fx) lt2.delet(fy);
if(rt == fy) lt2.delet(fx);
}
}
if(op[0] == 'A') {
if(op[1] == '1') {
x = readint();
y = readint();
int fx = lt1.froot(x), rt;
lt2.delet(fx);
rt = lt1.addpoint(x, y);
lt2.val[rt] = lt1.val[rt];
lt2.clear(rt);
lt2.root = lt2.merge(lt2.root, rt);
}
if(op[1] == '2') {
x = readint();
y = readint();
int fx = lt1.froot(x);
lt2.delet(fx);
lt1.val[fx] += y;
lt1.add[fx] += y;
lt2.val[fx] = lt1.val[fx];
lt2.clear(fx);
lt2.root = lt2.merge(lt2.root, fx);
}
if(op[1] == '3') {
x = readint();
add += x;
}
}
if(op[0] == 'F') {
if(op[1] == '1') {
x = readint();
printf("%d\n", lt1.val[x] + lt1.toadd(x) + add);
}
if(op[1] == '2') {
x = readint();
printf("%d\n", lt1.val[lt1.froot(x)] + add);
}
if(op[1] == '3') {
printf("%d\n", lt2.val[lt2.root] + add);
}
}
}
return 0;
}
调试留下来的数据生成器
因为经常WA,手写了一个数据生成器,偶尔会崩掉不要介意qwq,可自由调整n和q的取值。
// Code by KSkun, 2018/2
#include <cstdio>
#include <cstdlib>
#include <ctime>
int n, q;
char str[10][5] = {"U", "A1", "A2", "A3", "F1", "F2", "F3"};
int main() {
freopen("p3273.in", "w", stdout);
srand(time(NULL));
n = rand() % 50;
q = rand() % 50;
printf("%d\n", n);
for(int i = 1; i <= n; i++) {
printf("%d ", rand() - RAND_MAX / 2);
}
printf("\n");
printf("%d\n", q);
for(int i = 1; i <= q; i++) {
int op = rand() % 7;
printf("%s ", str[op]);
if(op == 0) {
printf("%d %d ", rand() % n + 1, rand() % n + 1);
}
if(op == 1 || op == 2 || op == 4 || op == 5) {
printf("%d ", rand() % n + 1);
}
if(op == 1 || op == 2 || op == 3) {
printf("%d ", rand() - RAND_MAX / 2);
}
printf("\n");
}
return 0;
}