P5284 [十二省联考2019]字符串问题 题解
SAM + 拓扑排序
主要说一下特殊的建图方式，自己脑补了一下，发现和题解不一样
首先，建出反串 SAM，然后把每个 \(a_i\) 挂到对应的 SAM 节点上，按照长度从小到大排序
然后把每个 SAM 节点拆成 2 个 \(s, t\)，\(s\) 向当前节点上挂的第一个 \(a_i\) 连边，边权是 \(|a_i|\)（长度），\(a_i\) 向后面的节点连边，边权是 \(|a_{i + 1}| - |a_{i}|\)（长度差）,最后一个节点 \(a_k\) 向 \(t\) 连边权是 \(-|a_k|\) 的边，然后 \(t\) 向原来的这个节点的儿子连边权是 0 的边
如果 \(a_i\) 能控制 \(b_j\) 直接从 \(a_i\) 向 \(b_j\) 连边，权值为 0
找出每个 \(b_i\) 对应的节点，二分一下，找到长度 \(\ge |b_i|\) 的第一个点 \(a_i\)，向它连边权为 \(|a_i|\) 的边
最后拓扑排序找最长路就行了
#include <bits/stdc++.h>
using namespace std;
#define gc getchar
#define rg register
#define I inline
#define rep(i, a, b) for(int i = a; i <= b; ++i)
#define per(i, a, b) for(int i = a; i >= b; --i)
I int read(){
rg char ch = gc();
rg int x = 0, f = 0;
while(!isdigit(ch)) f |= (ch == '-'), ch = gc();
while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc();
return f ? -x : x;
}
const int N = 4e5 + 5, M = N << 2;
int tr[N][26], len[N], lst, link[N], cnt, star[N >> 1], dfn[N], bk[N];
int fa[N][19], n, na, nb, pos[N];
int head[M], ver[M << 1], nxt[M << 1], tot, edge[M << 1], deg[M];
long long f[M];
int base;
vector<int> g[N << 1];
void add(int x, int y, int z){
swap(x, y);
ver[++tot] = y;
nxt[tot] = head[x];
head[x] = tot;
edge[tot] = z;
++deg[y];
//cout << y << " " << x << " " << z << endl;
}
char s[N];
struct node{
int l, r, id;
inline bool operator < (const node &rhs) const {
return r - l < rhs.r - rhs.l;
}
node(){}
node(int l, int r, int id): l(l), r(r), id(id) {}
};
vector<node> h[N];
I void prepro(){
rep(i, 0, base) head[i] = 0, deg[i] = 0, f[i] = 0;
tot = 0; base = 0;
rep(i, 0, cnt){
vector<node>().swap(h[i]);
memset(tr[i], 0, sizeof tr[i]);
vector<int>().swap(g[i]);
}
cnt = lst = 1;
}
I void insert(int c, int id){
int p = lst, cur = ++cnt; lst = cur;
len[cur] = len[p] + 1; star[id] = cur;
while(p && !tr[p][c]) tr[p][c] = cur, p = link[p];
if(!p) link[cur] = 1;
else{
int q = tr[p][c];
if(len[q] == len[p] + 1) link[cur] = q;
else{
int clone = ++cnt;
len[clone] = len[p] + 1; link[clone] = link[q]; memcpy(tr[clone], tr[q], sizeof tr[q]);
while(p && tr[p][c] == q) tr[p][c] = clone, p = link[p];
link[cur] = link[q] = clone;
}
}
}
void dfs(int x){
rep(i, 1, 18) fa[x][i] = fa[fa[x][i - 1]][i - 1];
for(int y : g[x]) fa[y][0] = x, dfs(y);
}
I void build(){
per(i, n, 1) insert(s[i] - 'a', i);
//rep(i, 2, cnt) cout << link[i] << i << endl;
rep(i, 2, cnt) g[link[i]].push_back(i);
dfs(1);
}
I int getfa(int l, int r){
int L = r - l + 1;
int x = star[l];
for(int i = 18; ~i; --i){
if(L <= len[fa[x][i]]) x = fa[x][i];
}
return x;
}
I long long topu(){
queue<int> q;
rep(i, 1, base) if(!deg[i]) q.push(i);
int num = 0;
while(!q.empty()){
int x = q.front(); q.pop();
++num;
for(int i = head[x]; i; i = nxt[i]){
int y = ver[i];
f[y] = max(f[y], f[x] + edge[i]);
--deg[y];
if(!deg[y]) q.push(y);
}
}
if(num != base) return -1;
return f[1];
}
I void Main(){
prepro();
scanf("%s", s + 1); n = strlen(s + 1);
build();
na = read();
int l, r;
rep(i, 1, na){
l = read(), r = read();
int t = getfa(l, r);
h[t].push_back(node(l, r, i));
}
base = cnt;
rep(i, 1, cnt){
bk[i] = base + na + i;
for(int y : g[i]) add(bk[i], y, 0);
sort(h[i].begin(), h[i].end());
if(h[i].begin() != h[i].end()){
add(i, base + h[i].begin() -> id, h[i].begin() -> r - h[i].begin() -> l + 1);
add(base + (h[i].end() - 1) -> id, bk[i], (h[i].end() - 1) -> l - (h[i].end() - 1) -> r - 1);
for(vector<node>::iterator it = h[i].begin(); it != h[i].end(); ++it)
if((it + 1) != h[i].end()) add(it -> id + base, (it + 1) -> id + base, (it + 1) -> r - (it + 1) -> l - it -> r + it -> l);
}else add(i, bk[i], 0);
}
nb = read();
rep(i, 1, nb){
l = read(), r = read();
pos[i] = getfa(l, r); vector<node>::iterator it = lower_bound(h[pos[i]].begin(), h[pos[i]].end(), (node){l, r, i});
if(it == h[pos[i]].end()) add(base + na + cnt + i, bk[pos[i]], 0);
else add(base + na + cnt + i, base + it -> id, it -> r - it -> l + 1);
}
base += na + nb + cnt;
int m = read();
rep(i, 1, m){
l = read(), r = read();
add(cnt + l, cnt + na + cnt + r, 0);
}
printf("%lld\n", topu());
}
signed main(){
int T = read();
while(T--) Main();
return 0;
}
