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题目链接
https://vjudge.net/contest/248768#problem/M
题目大意
有n 个栈, 每个栈有 个元素 , 每次只能依次某个栈的栈顶取出一个元素, 获得的费用为 , D表示该元素是倒数第D个取的。 最大化费用, 输出 mod + 7
题目思路
本质是每次选择某个数位的值, 最大化一个365进制的数, 很容易想到贪心, 很明显是优先最高位最小, 每次从所有的栈顶中选取最小的值, 再依次出栈, 弄一个堆即可完成。 但问题是如果出现了相等的值该如何考虑呢。
显然随便选一个肯定是不对的, 还要往后一直比较下去。
实际上就是在比较两个后缀的字典序。
联想到后缀数组可以解决, 将各个栈的元素拼接成一个串, 中间用一个最大值301分割, 求一遍后缀数组, 堆中比较的是对应后缀的后缀排名。
Code
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <bitset>
#include <map>
#include <stack>
#include <set>
#define ls ch[x][0]
#define rs ch[x][1]
#define ll long long
#define pi pair<int, int>
#define mp make_pair
#define fi first
#define se second
using namespace std;
int gi(){
int ret = 0; char c = getchar();
while (!isdigit(c)) c = getchar();
while (isdigit(c)){
ret = ret * 10 + c - '0';
c = getchar();
}
return ret;
}
const int N = (int)1e6 + 10;
const int mo = (int)1e9 + 7;
int n, m;
int str[N], st[N], ed[N], sz[N]; ll pw[N];
int sa[N], rk[N], c[N], x[N], y[N];
void build_sa(){
m = 310;
for (int i = 0; i <= m; i ++) c[i] = 0;
for (int i = 1; i <= ed[n]; i ++) x[i] = str[i];
for (int i = 1; i <= ed[n]; i ++) c[x[i]] ++;
for (int i = 2; i <= m; i ++) c[i] += c[i - 1];
for (int i = 1; i <= ed[n]; i ++) sa[c[x[i]] --] = i;
for (int i = 1; i <= m; i ++) c[i] = 0;
for (int k = 1; k < ed[n]; k <<= 1){
int p = 0;
for (int i = ed[n] - k + 1; i <= ed[n]; i ++) y[++ p] = i;
for (int i = 1; i <= ed[n]; i ++) if (sa[i] > k) y[++ p] = sa[i] - k;
for (int i = 1; i <= ed[n]; i ++) c[x[i]] ++;
for (int i = 2; i <= m; i ++) c[i] += c[i - 1];
for (int i = ed[n]; i >= 1; i --) sa[c[x[y[i]]] --] = y[i];
for (int i = 1; i <= m; i ++) c[i] = 0;
for (int i = 1; i <= ed[n]; i ++) y[i] = x[i];
p = 1;
x[sa[1]] = 1;
for (int i = 2; i <= ed[n]; i ++)
x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p : ++p;
for (int i = 1; i <= ed[n]; i ++) y[i] = 0;
if (p >= ed[n]) break;
m = p;
}
}
int main(){
pw[0] = 1;
for (int i = 1; i < N; i ++) pw[i] = pw[i - 1] * 365 % mo;
while (scanf("%d", &n) != EOF){
int T = 0;
for (int i = 1; i <= n; i ++){
sz[i] = gi();
st[i] = ed[i - 1] + 1;
for (int j = 0, x; j < sz[i]; j ++){
x = gi(); str[st[i] + j] = x;
}
str[st[i] + sz[i]] = 301;
ed[i] = st[i] + sz[i];
T += sz[i];
}
build_sa();
for (int i = 1; i <= ed[n]; i ++) rk[sa[i]] = i;
priority_queue<pi > H;
for (int i = 1; i <= n; i ++){
H.push(mp(-rk[st[i]], i));
st[i] ++;
}
ll ans = 0;
while (H.size()){
int i = H.top().se;
H.pop();
ans = (ans + str[st[i] - 1] * pw[T --] % mo) % mo;
if (st[i] < ed[i]){
H.push(mp(-rk[st[i]], i));
st[i] ++;
}
}
printf("%lld\n", ans);
}
return 0;
}