#6：年兽礼包——6

石子合并，入门区间dp

#include <cstdio>
#include <cstring>
#define maxn 305
#define min(a, b) a < b ? a : b

int n;
int sum[maxn];
int f[maxn][maxn];

int main() {
    memset(f, 0x3f, sizeof(f));
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &sum[i]);
        sum[i] += sum[i-1]; 
        f[i][i] = 0;
    }
    for (int len = 2; len <= n; len++)
        for (int l = 1; l <= n-len+1; l++) {
            int r = l + len - 1;
            for (int k = l; k < r; k++)
                f[l][r] = min(f[l][r], f[l][k] + f[k+1][r]);
            f[l][r] += sum[r] - sum[l-1];
        }
    printf("%d\n", f[1][n]);
    return 0;
}

 

BZOJ3229，还是石子合并，但是数据量大，换算法了。100ms以内的那个实现真是厉害！我这个是朴素的……

#include <cstdio>
#define maxn 40005
#define INF 0x7fffffff
#define R(a) scanf("%d", &a)
#define swap(a, b) a^=b^=a^=b
#define max(a, b) a > b ? a : b

int n, a[maxn], ans, k, st = 1;

int main() {
    R(n);
    for (int i = 1; i <= n; i++)
        R(a[i]);
    a[0] = a[n+1] = INF;

    for (int m = n, i = 1; i < n; i++, m--) {
        for (int j = st; j <= m; j++)
            if (a[j-1] <= a[j+1]) {
                k = j;
                break;
            }
        
        a[k-1] += a[k];
        ans += a[k-1];

        for (int j = k; j <= m; j++)
            a[j] = a[j+1];

        st = k;
        for(k--; a[k-1] <= a[k]; swap(a[k-1], a[k]), k--);

        if (k > 2 && a[k-2] <= a[k])    st = k-1;
        else    st = max(st-2, 1);
    }

    printf("%d\n", ans);
    return 0;
}

 

POJ1179，环形拆链，乘法有负数没法直接dp，石子合并的基础上再维护一个最小值即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int n, ans = 0xcfcfcfcf;
int Digit[105];
bool signal[105];
int f[105][105], g[105][105];
int record[55], t;

inline int cal(int a, int b, int k) {
    return signal[k] ? a+b : a*b;
}

int main() {
    memset(f, 0xcf, sizeof(f));
    memset(g, 0x3f, sizeof(g));
    cin >> n;
    for (int i = 1; i <= n; i++) {
        char ch[3];
        scanf("%s%d", ch, &Digit[i]);
        signal[i-1] = signal[i-1+n] = ch[0] == 't' ? true : false;
        f[i][i] = f[i+n][i+n] = g[i][i] = g[i+n][i+n] = Digit[i+n] = Digit[i];
    }

    for (int len = 2; len <= n; len++)
        for (int l = 1; l <= 2*n - len + 1; l++) {
            int r = l + len - 1;
            for (int k = l; k < r; k++) {
                f[l][r] = max(f[l][r], cal(f[l][k], f[k+1][r], k));
                g[l][r] = min(g[l][r], cal(g[l][k], g[k+1][r], k));
                if (!signal[k])    {
                    f[l][r] = max(f[l][r], cal(g[l][k], g[k+1][r], k));
                    g[l][r] = min(g[l][r], min(cal(f[l][k], g[k+1][r], k), cal(g[l][k], f[k+1][r], k)));
                }
            }
        }
    
    for (int i = 1; i <= n; i++)
        if (ans <= f[i][i+n-1]) {
            if (ans < f[i][i+n-1])    t = 0;
            ans = f[i][i+n-1];
            record[++t] = i;
        }

    cout << ans << endl;
    for (int i = 1; i <= t; i++)
        printf("%d%c", record[i], " \n"[i==t]);

    return 0;
}

 

CH金字塔，将dfs序和区间联系起来。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define maxn 305
#define mod 1000000000
using namespace std;

char S[maxn];
ll f[maxn][maxn];

int solve(int l, int r) {
    if (~f[l][r])    return f[l][r];
    if (l == r)    return f[l][r] = 1;
    if (S[l-1] != S[r-1])    return f[l][r] = 0;
    
    f[l][r] = 0;
    for (int k = l+1; k < r; k++)
        f[l][r] = (f[l][r] + (ll)solve(l+1, k) * solve(k+1, r)) % mod;
    return f[l][r];
}

int main() {
    cin >> S;
    memset(f, -1, sizeof(f));
    printf("%d\n", solve(1, strlen(S)));
    return 0;
}

 

Joyoi没有上司的舞会，入门树形dp。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#define maxn 6005
#define pb push_back
using namespace std;

int n, H[maxn], f[maxn];
int m[maxn][2];
vector<int> v[maxn];

int dp(int cur,int select) {
    if (m[cur][select] != 0xcfcfcfcf)    return m[cur][select];

    if (select) {
        m[cur][select] = H[cur];
        for (auto i : v[cur])
            m[cur][select] += dp(i, 0);
    } else {
        m[cur][select] = 0;
        for (auto i : v[cur])
            m[cur][select] += max(dp(i, 0), dp(i, 1));
    }

    return m[cur][select];
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> H[i];

    int so, fa;
    while (cin >> so >> fa && so) {
        f[so] = fa;
        v[fa].pb(so);
    }
    for (fa = 1; f[fa]; fa = f[fa]);

    memset(m, 0xcf, sizeof(m));
    printf("%d\n", max(dp(fa, 1), dp(fa, 0)));
    return 0;
}

 

Joyoi选课，树上分组背包，果然j从0~t也是对的，没太懂书上非要倒序的意思qwq

#include <bits/stdc++.h>
#define maxn 305
#define pb push_back
using namespace std;

int n, m, score[maxn];
int f[maxn][maxn];
std::vector<int> v[maxn];

int dp(int cur) {
    f[cur][0] = 0;
    for (auto i : v[cur]) {
        dp(i);
        for (int t = m; ~t; t--)
            for (int j = 0; j <= t; j++)
                f[cur][t] = max(f[cur][t], f[cur][t-j] + f[i][j]);
    }
    if (cur)
        for (int t = m; t; t--)
            f[cur][t] = f[cur][t-1] + score[cur];
}

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        int a;
        cin >> a >> score[i];
        v[a].pb(i);
    }
    memset(f, 0xcf, sizeof(f));
    dp(0);
    printf("%d\n", f[0][m]);
    return 0;
}