bzoj1563: [NOI2009]诗人小G 决策单调性(1D1D)

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目录

• [TOC]

题目链接

bzoj1563: [NOI2009]诗人小G

题解

\(n^2\) 的dp长这样

\(f_i = min(f_j + (sum_i - sum_j - 1 - L)^P)\)

设\(w_{ij} = (sum_i - sum_j - 1 - L)^P\)

那么化成1D1D的标准形式

$ f_i = min(f_j + w_{i,j}) $

发现w满足四边形不等式

证明可以看这里

https://www.byvoid.com/zhs/blog/noi-2009-poet

因此状态转移方程具有单调性

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define gc getchar()
#define pc putchar
#define LD long double
inline int read() {
    int x = 0,f = 1;
    char c = gc;
    while(c < '0' || c > '9' )c = gc;
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
    return x * f ;
}
void print(LL x) {
    if(x >= 10) print(x / 10);
    pc(x % 10 + '0');
}
const int maxn = 100007;
char s[maxn][32];
int n,L,p;
inline LD fstpow(LD x,int k) {
    LD ret = 1;
    for(;k;k >>= 1,x = x * x) if(k & 1) ret *= x;
    return ret;
}
LD f[maxn];
int sum[maxn];
LD calc(int j,int i) {
    return f[j] + fstpow(std::abs(sum[i] - sum[j] - L),p);
}
int find(int x,int y) {
    int l = x,r = n,ret = 0;
    while(l <= r) {
        int mid = l + r >> 1;
        if(calc(x,mid) >= calc(y,mid)) r = mid - 1;
        else l = mid + 1;
    }
    return l;
}
int q[maxn],c[maxn];
int pre[maxn];
void solve() {
    n = read(),L = read() + 1,p = read();
    for(int i = 1;i <= n;++ i) {
        scanf("%s",s[i] + 1);
        sum[i] = sum[i - 1] + strlen(s[i] + 1) + 1;
    }
    int h = 1,t = 1;
    q[h] = 0;
    for(int i = 1;i <= n;++ i) {
        while(h < t && c[h] <= i) ++ h;
        f[i] = calc(q[h],i); pre[i] = q[h];
        while(h < t && c[t - 1] >= find(q[t],i)) t --;
        c[t] = find(q[t],i); q[++ t] = i;
    }
    if(f[n] > 1e18) {
        puts("Too hard to arrange\n--------------------");
        return;
    }
    printf("%.0Lf\n", f[n]); 

    puts("--------------------");
}
int main() {
    int T = read();
    for(int i = 1; i <= T; ++ i) {
        solve();
    }
    return 0;
}