[BZOJ 4380] POI 2015 Myjnie

来源：互联网发布：用c语言写出九九乘法表编辑：程序博客网时间：2024/05/14 21:57

这里写图片描述

网上搜不到题解…搞了两天才勉强跑过去…太弱…

做法应该就是区间dp吧，把所有询问的 c 值排序，由大到小枚举每一个 c ，令 f[i][j]表示当前的区间 [i, j] 中能取到的最大值，然后用一种神乎其神的方法状态转移和记录答案…具体可以看代码.

AC code :

#include<cstdio>#include<algorithm>using namespace std;typedef long long LL;inline void read(int &x){x=0;char c;while((c=getchar())<'0'||c>'9');for(x=c-'0';(c=getchar())>='0'&&c<='9';x=x*10+c-'0');}const int N = 51, M = 5001 ;struct Query{    int l, r, m;}q[M];int f[N][N], g[M][N][N];int T[M], n, m, c;struct Data{    Data *ch[2];    int p, k;}P[N * (N + 1) * M / 2], *head[N][N], *cur = P;inline void insert(int l, int r, int d, int k){    cur->ch[0] = head[l][d - 1], cur->ch[1] = head[d + 1][r];    cur->p = d, cur->k = k, head[l][r] = cur++;}void dfs(Data *o, int l, int r){    int d = o->p;    if (l < d) dfs(o->ch[0], l, d - 1);    printf("%d ", T[o->k]);;    if (d < r) dfs(o->ch[1], d + 1, r);}int main(){    freopen("in.txt", "r", stdin);    read(n), read(m);    for (int i = 1; i <= m; ++i)        read(q[i].l), read(q[i].r), read(q[i].m), T[i] = q[i].m;    sort(T + 1, T + m + 1);    c = unique(T + 1, T + m + 1) - T - 1;    for (int i = 1; i <= m; ++i)    {        q[i].m = lower_bound(T + 1, T + c + 1, q[i].m) - T;        int l = q[i].l, r = q[i].r;        for (int k = 1; k <= q[i].m; ++k) g[k][l][r] += T[k];    }    for (int k = 1; k <= c; ++k) for (int l = n; l; --l) for (int r = l; r <= n; ++r)        g[k][l][r] += g[k][l + 1][r] + g[k][l][r - 1] - g[k][l + 1][r - 1];    for (int i = 1; i <= n; ++i) for (int j = i; j <= n; ++j) f[i][j] = -1;    for (int k = c; k; --k)    {        for (int l = 0; l < n; ++l) for (int i = 1; i + l <= n; ++i)        {            int j = i + l, ret = 0, p = i;            for (int d = i; d <= j; ++d)            {                int tmp = f[i][d - 1] + f[d + 1][j] + g[k][i][j] - g[k][i][d - 1] - g[k][d + 1][j];                if (ret < tmp) ret = tmp, p = d;            }            if (f[i][j] < ret) f[i][j] = ret, insert(i, j, p, k);        }    }    printf("%d\n", f[1][n]);    dfs(head[1][n], 1, n);    putchar('\n');    return 0;}

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