hdu 1069 Monkey and Banana

題目鏈接

其實，dp我一直是拒絕的。這方面我一直十分薄弱。但是難關還是一定要攻克的＝ ＝如是我終於開始磕dp專題了。那就先從比較簡單的問題開始吧。

題目大意是有一些不同規格立方塊無限個，把它們疊起來。要求下方的在橫截面上長寬嚴格大於上面的，問疊起來最高的高度。

分析：

考慮到立方塊可以交換長寬高，所以我把一個立方塊變成六個不同的立方塊來思考。將變換長寬高後的立方塊按照關鍵詞先後先a從小到大，再b從大到小，最後c從小到大的順序排序並且去重。這樣，我們能夠簡化考慮的維數。因爲a已經按照從小到大的順序排好了，我們只需要考慮b的大小就可以了。不難想到一個遞推式子dp[i]=max(dp[k])+height[i] 其中要求第k個立方塊的b[k]

代碼：

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)
#define   DEBUG         freopen("#.txt", "r", stdin)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   ll            long long
#define   ull           unsigned long long
#define   lowbit(x)     (x&-x)

const int    INF    = 0x3f3f3f3f;
const ll     INFF   = 1e18;
const double pi     = acos(-1.0);
const double inf    = 1e18;
const double eps    = 1e-9;
const ll     mod    = 1e9+7;
const int    maxmat = 10;
const ull    BASE   = 133333331;

/*****************************************************/
inline void RI(int &x) {
      char c;
      while((c=getchar())<'0' || c>'9');
      x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
inline ll bits(ll x) {
      return !x ? x : bits(x - lowbit(x)) + 1;
}
/*****************************************************/

const int maxn = 35;

struct Node {
    int a, b, c;
    bool operator <(const Node &rhs) const {
        if (a == rhs.a && b == rhs.b) return c < rhs.c;
        if (a == rhs.a) return b > rhs.b;
        return a < rhs.a;
    }
    bool operator ==(const Node &rhs) const {
        return a == rhs.a && b == rhs.b && c == rhs.c;
    }
}node[maxn * 6];

int N;
int lis[maxn * 3];
int tree[(maxn * 6) << 2];
int dp[maxn * 6];

void PushUp(int v) {
    tree[v] = max(tree[slch], tree[srch]);
}

void update(int x, int val, int l, int r, int v) {
    if (l == r) tree[v] = max(tree[v], val);
    else {
        int m = (l + r) >> 1;
        if (x <= m) update(x, val, lson);
        else update(x, val, rson);
        PushUp(v);
    }
}

int query(int L, int R, int l, int r, int v) {
    if (L > R) return 0;
    if (L <= l && r <= R) return tree[v];
    int m = (l + r) >> 1;
    int ans = -1;
    if (L <= m) ans = max(ans, query(L, R, lson));
    if (R >  m) ans = max(ans, query(L, R, rson));
    return ans;
}

int main(int argc, char const *argv[]) {
    int kase = 1;
    while (~scanf("%d", &N) && N) {
        memset(tree, 0, sizeof(tree));
        int tot = 0, cnt = 0;
        for (int i = 0; i < N; i ++) {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            node[tot ++] = (Node){a, b, c};
            node[tot ++] = (Node){b, a, c};
            node[tot ++] = (Node){a, c, b};
            node[tot ++] = (Node){c, a, b};
            node[tot ++] = (Node){b, c, a};
            node[tot ++] = (Node){c, b, a};
            lis[cnt ++] = a;
            lis[cnt ++] = b;
            lis[cnt ++] = c;
        }

        sort(node, node + tot);
        sort(lis, lis + cnt);
        int len = unique(node, node + tot) - node;
        int num = unique(lis, lis + cnt) - lis;

        for (int i = 0; i < len; i ++) {
            int p = lower_bound(lis, lis + num, node[i].b) - lis;
            int d = query(0, p - 1, 0, num, 1);
            dp[i] = d + node[i].c;
            update(p, dp[i], 0, num, 1);
        }

        printf("Case %d: maximum height = %d\n", kase ++, *max_element(dp, dp + len));
    }
    return 0;
}