HDU 5900 QSC and Master Sheng Yang 网络赛

来源：互联网发布：mac最新系统下载地址编辑：程序博客网时间：2024/06/07 01:06

题解：我们可以先通过dp判断哪些可以进行合并然后再通过一维dp 得到最大值

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#define ll long longusing namespace  std;const int maxn = 505;ll d[maxn] = {0};bool dp[maxn][maxn] = {0};ll a[maxn] = {0};ll b[maxn] = {0};ll sum[maxn] = {0};ll ans = 0;int main () {    ios_base :: sync_with_stdio(false);    int t;    cin >> t;    while (t--) {        int n;        cin >> n;        memset (d,0,sizeof (d));        memset (dp,0,sizeof (dp));        memset (sum,0,sizeof (sum));        for (int i = 1;i <= n; ++ i) cin >> a[i];        for (int i = 1;i <= n; ++ i){            cin >> b[i];            sum[i] += b[i];            sum[i] += sum[i - 1];        }        for (int i = 1;i < n; ++ i) {            if (__gcd(a[i],a[i + 1]) != 1) {                dp[i][i + 1] = 1;            }        }        for (int l = 3;l <= n; l += 2) {            for (int i = 1;i + l <= n; ++i) {                int j = i + l;                for (int k = i + 1;k < j; ++ k) {                    if (dp[i][k] && dp[k + 1][j]) {                        dp[i][j] = 1;                    }                }                if (__gcd(a[i],a[j]) != 1 && dp[i + 1][j - 1]) {                    dp[i][j] = 1;                }            }        }        for (int i = 1;i <= n; ++ i) {            d[i] = d[i - 1];            for (int j = 1;j < i; ++ j) {                if (dp[j][i]) {                    d[i] = max (d[i],d[j - 1] + sum[i] - sum[j - 1]);                }            }        }//        cout << dp[1][n] << endl;        cout << d[n] << endl;    }    return 0;}

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