bzoj刷题记录5.11-5.15

bzoj2809:[Apio2012]dispatching

重写个左偏树版的。
每次把不合法的删去，留下剩下的，然后合并上去，总复杂度O(nlgn)。

#include 
using namespace std;

const int MAXN = 100005;

struct ltheap {
    struct node {
        long long dat;
        int lc, rc, dist;
    } tree[MAXN*4];
    int top;
    ltheap(){ top = 0; }
    int new_node(long long dat)
    { return tree[++top].dat = dat, tree[top].dist = 0, top; }
    int merge(int &a, int &b)
    {
        if (!a && !b) return 0;
        if (!a) { a = b; return a; }
        if (tree[a].dat < tree[b].dat) swap(a, b);
        merge(tree[a].rc, b);
        if (tree[tree[a].lc].dist < tree[tree[a].rc].dist)
            swap(tree[a].lc, tree[a].rc);
        tree[a].dist = tree[tree[a].rc].dist + (tree[a].rc != 0);
        return a;
    }
    long long pop_top(int nd)
    {
        long long ret = tree[nd].dat;
        tree[nd] = tree[merge(tree[nd].lc, tree[nd].rc)];
        return ret;
    }
    void touring(int nd, int tab)
    {
        if (!nd) return;
        for (int i = 1; i <= tab; i++) putchar(' ');
        printf("%lld, dist = %d\n", tree[nd].dat, tree[nd].dist);
        touring(tree[nd].lc, tab+2);
        touring(tree[nd].rc, tab+2);
    }
} ltheap;

long long c[MAXN], l[MAXN], ans = 0;
int lab[MAXN], num[MAXN];
long long cnt[MAXN], M;

struct node {
    int to, next;
} edge[MAXN];
int head[MAXN], top = 0;
void push(int i, int j)
{ edge[++top] = (node){j, head[i]}, head[i] = top; }

int n, rt;

void dfs(int nd)
{
    lab[nd] = ltheap.new_node(c[nd]), cnt[nd] += c[nd], num[nd] = 1;
    for (int i = head[nd]; i; i = edge[i].next) {
        int to = edge[i].to;
        // cout << to << endl;
        dfs(to), lab[nd] = ltheap.merge(lab[nd], lab[to]);
        cnt[nd] += cnt[to], num[nd] += num[to];
    }
    while (cnt[nd] > M)
        cnt[nd] -= ltheap.pop_top(lab[nd]), num[nd]--;
    // cout << nd << " " << num[nd] << " " << cnt[nd] << " " << l[nd] << endl;
    ans = max(ans, num[nd]*l[nd]);
}

int main()
{
    scanf("%d%lld", &n, &M);
    for (int i = 1; i <= n; i++) {
        int u;
        scanf("%d%lld%lld", &u, &c[i], &l[i]);
        if (u == 0) rt = i;
        else push(u, i);
    }
    dfs(rt);
    cout << ans << endl;
    return 0;
}

bzoj1296: [SCOI2009]粉刷匠

sb题啊就是…脑残了半天

#include 
using namespace std;

const int MAXN = 55;
char str[MAXN];
int dat[MAXN][MAXN], T;
int dp[MAXN][MAXN][MAXN]; // 第i个，前j个，k次
int s[MAXN];
int f[MAXN][MAXN*MAXN], n, m;

int main()
{
    scanf("%d%d%d", &n, &m, &T);
    for (int i = 1; i <= n; i++) {
        scanf("%s", str+1);
        for (int j = 1; j <= m; j++)
            dat[i][j] = str[j]-'0';
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) s[j] = s[j-1]+dat[i][j];
        for (int j = 1; j <= m; j++)
            for (int k = 1; k <= m; k++)
                for (int l = 0; l < j; l++)
                    dp[i][j][k] = max(dp[i][j][k], dp[i][l][k-1]+max(s[j]-s[l], j-l-(s[j]-s[l])));
    }
    for (int i = 1; i <= n; i++) // dp[i][m][Times]
        for (int j = 0; j <= T; j++) {
            for (int k = 0; k <= min(j, m); k++)
                f[i][j] = max(f[i][j], f[i-1][j-k]+dp[i][m][k]);
        }
    cout << f[n][T] << endl;
    return 0;
}

bzoj3233: [Ahoi2013]找硬币

状态设计极其诡异的dp…

首先一个性质是：如果能用大额支付一定不会用小额..这很显然，因此如果用大额k×i代替i来支付a[j]价格可以多用大额票票：

?a[j]i?k

张代替小额票票

i?a[j]i?

张。于是就有了方程：

dp[ik]=min?????dp[ik],dp[i]?∑j(i?a[j]i?+?a[j]i?k)?????

由于∑i≤nφ(i)=O(nlnn)，总复杂度是O(nmlnn)的，可以过掉。

#include 
using namespace std;

const int MAXN = 55, MAXA = 100005;

int f[MAXA], n;
int a[MAXN], ma;

int main()
{
    scanf("%d", &n);
    memset(f, 127, sizeof f);
    f[1] = ma = 0;
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]), f[1] += a[i], ma = max(ma, a[i]);
    for (int i = 1; i <= ma; i++)
        for (int j = 2; j*i <= ma; j++) {
            int ans = f[i];
            for (int k = 1; k <= n; k++) {
                int tms = a[k]/i/j;
                ans += tms-tms*j;
            }
            f[j*i] = min(f[j*i], ans);
        }
    int ans = 233333333;
    for (int i = 1; i <= ma; i++) {
        ans = min(ans, f[i]);
    }
    cout << ans << endl;
    return 0;
}