zoukankan      html  css  js  c++  java
  • Blocks poj 区间dp

    Some of you may have played a game called 'Blocks'. There are n blocks in a row, each box has a color. Here is an example: Gold, Silver, Silver, Silver, Silver, Bronze, Bronze, Bronze, Gold.
    The corresponding picture will be as shown below:

    Figure 1

    If some adjacent boxes are all of the same color, and both the box to its left(if it exists) and its right(if it exists) are of some other color, we call it a 'box segment'. There are 4 box segments. That is: gold, silver, bronze, gold. There are 1, 4, 3, 1 box(es) in the segments respectively.

    Every time, you can click a box, then the whole segment containing that box DISAPPEARS. If that segment is composed of k boxes, you will get k*k points. for example, if you click on a silver box, the silver segment disappears, you got 4*4=16 points.

    Now let's look at the picture below:

    Figure 2


    The first one is OPTIMAL.

    Find the highest score you can get, given an initial state of this game.
    Input
    The first line contains the number of tests t(1<=t<=15). Each case contains two lines. The first line contains an integer n(1<=n<=200), the number of boxes. The second line contains n integers, representing the colors of each box. The integers are in the range 1~n.
    Output
    For each test case, print the case number and the highest possible score.
    Sample Input
    2
    9
    1 2 2 2 2 3 3 3 1
    1
    1
    Sample Output
    Case 1: 29
    Case 2: 1

    对于贪心显然就不正确了;
    那么考虑dp;
    设dp[ i ][ j ][ k ]表示i~j区间,最后合并k个的最大值;

    dp[ i ][ j ][ k ]=dp[ i ][ j-1 ][ 0 ]+( len[ j ]+k )^2;
    第二种情况就是中间一段先消去,然后与后面那一段拼接消除;
    dp[ i ][ j ][ k ]=dp[ i ][ k ][ len[ j ]+k ]+dp[ k+1 ][ j-1 ][ 0 ];
    那么我们记忆化dfs即可;
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 400005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    const long long int mod = 1e9 + 7;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-3
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    inline ll rd() {
    	ll x = 0;
    	char c = getchar();
    	bool f = false;
    	while (!isdigit(c)) {
    		if (c == '-') f = true;
    		c = getchar();
    	}
    	while (isdigit(c)) {
    		x = (x << 1) + (x << 3) + (c ^ 48);
    		c = getchar();
    	}
    	return f ? -x : x;
    }
    
    ll gcd(ll a, ll b) {
    	return b == 0 ? a : gcd(b, a%b);
    }
    ll sqr(ll x) { return x * x; }
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
    	if (!b) {
    		x = 1; y = 0; return a;
    	}
    	ans = exgcd(b, a%b, x, y);
    	ll t = x; x = y; y = t - a / b * y;
    	return ans;
    }
    */
    
    ll mode;
    struct matrix {
    	ll n, m, a[10][10];
    	matrix(ll n, ll m) {
    		this->n = n; this->m = m; ms(a);
    	}
    	matrix(ll n, ll m, char c) {
    		this->n = n; this->m = m; ms(a);
    		for (int i = 1; i <= n; i++)a[i][i] = 1;
    	}
    	ll *operator [](const ll x) {
    		return a[x];
    	}
    	matrix operator *(matrix b) {
    		matrix c(n, b.m);
    		for (int i = 1; i <= n; i++) {
    			for (int j = 1; j <= b.m; j++) {
    				for (int k = 1; k <= m; k++) {
    					c[i][j] = (c[i][j] + a[i][k] % mode*b[k][j] % mode) % mode;
    				}
    			}
    		}
    		return c;
    	}
    	void operator *=(matrix &b) {
    		*this = *this *b;
    	}
    	matrix operator ^(ll b) {
    		matrix ans(n, m, 'e'), a = *this;
    		while (b) {
    			if (b % 2)ans = ans * a; a *= a; b >>= 1;
    		}
    		return ans;
    	}
    };
    
    
    int dp[202][202][202];
    int T;
    int n;
    int col[210];
    int len[202];
    int fg;
    int dfs(int x, int y, int k) {
    	if (dp[x][y][k])return dp[x][y][k];
    	if (x == y)return (len[x] + k)*(len[x] + k);
    	dp[x][y][k] = dfs(x, y - 1, 0) + (len[y] + k)*(len[y] + k);
    	for (int i = x; i < y; i++) {
    		if (col[i] == col[y]) {
    			dp[x][y][k] = max(dp[x][y][k], dfs(x, i, len[y] + k) + dfs(i + 1, y - 1, 0));
    		}
    	}
    	return dp[x][y][k];
    }
    
    int main()
    {
    	//ios::sync_with_stdio(0);
    	rdint(T); int cnt = 0;
    	while (T--) {
    		cnt++;
    		ms(dp); ms(col); ms(len);
    		fg = 0;
    		int ans = 0;
    		rdint(n);
    		for (int i = 1; i <= n; i++) {
    			int tmp; rdint(tmp);
    			if (col[fg] == tmp)len[fg]++;
    			else fg++, len[fg] = 1, col[fg] = tmp;
    		}
    		ans = dfs(1, fg, 0);
    		cout << "Case " << cnt << ": " << ans << endl;
    	}
    	return 0;
    }
    
    
    
    EPFL - Fighting
  • 相关阅读:
    [mysql] 5.1和5.5版本区别
    selenium-51job高级搜索
    selenium-xpath选择操作web元素
    selenium-css选择器高级用法
    selenium-51job自动化测试(css选择器选择元素)
    API测试
    接口测试
    selenium-百度新歌榜
    selenium-frame切换
    selenium等待元素出现和代码抛出异常继续执行
  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10187832.html
Copyright © 2011-2022 走看看