srm 552

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250

Description

定义大小为N的三角形, 是由若干个等大的圆形构成的, 高度和底宽为N,三角形的每个圆染三种颜色r,g,b,相接触的圆不能染同种颜色,问有R个r颜色的球, G个g颜色的球和B个b颜色的球, 最多能染多少个大小为N的三角形

Solution

稍加分析可以发现，每个三角形三个颜色的球要不是x,x,x,要不是x,x,x+1这种形式。先考虑mod3=0的情况，直接计算即可，mod3=1时，先当他们都相同，二分答案，然后check一下多出来的是否可以多出个数。

Code

#include <bits/stdc++.h>using namespace std;#define pb push_back#define mp make_pair#define F first#define S secondtypedef long long LL;typedef pair<int, int> pii;vector<LL> a;struct FoxPaintingBalls {    long long theMax(long long R, long long G, long long B, int N) {        if (N == 1) return R + G + B;        a.pb(R), a.pb(G), a.pb(B);        sort(a.begin(), a.end());        LL tot = (LL)N * (N + 1) / 2;        LL x = tot / 3;        if (tot % 3 == 0) return a[0] / x;        LL ll = 0, ans = 0, rr = a[0] / x;        while (ll <= rr) {            LL mid = (ll + rr) >> 1ll;            LL r = R - mid * x, g = G - mid * x, b = B - mid * x;            if (r + g + b >= mid) {                ans = mid;                ll = mid + 1;            }            else rr = mid - 1;        }        return ans;    }};

500

Description

大小为n×m(n,m<30)的矩阵, 有L,P,.,三种格子, 画两个互不相交的矩形, 使两个矩形L和P的差不超过D, 问这两个矩形最多能包含的L和P的和。

Solution

考虑一定存在一个分界线将其分为两个矩阵，先暴力N6处理所有分界线，比如l[i][j]表示第i列左边差为j的最大的花的和，然后枚举两个矩形内部的差，再枚举分界线，暴力N5计算即可。

Code

#include <bits/stdc++.h>//enumusing namespace std;#define pb push_back#define mp make_pair#define F first#define S secondtypedef long long LL;typedef pair<int, int> pii;const int N = 35;int L[N][N * N + 1000], R[N][N * N + 1000], D[N][N * N + 1000], U[N][N * N + 1000];inline void gao(int &x, int y) {    if (x < y)  x = y;}struct FoxAndFlowerShopDivOne {    int theMaxFlowers(vector <string> s, int maxDiff) {        memset(L, 0xc0, sizeof(L));        memset(R, 0xc0, sizeof(R));        memset(U, 0xc0, sizeof(U));        memset(D, 0xc0, sizeof(D));        int n = s.size(), m = s[0].size();        for (int i = 0; i < n; ++i)            for (int j = 0; j < m; ++j)                 for (int k = i; k < n; ++k)                    for (int l = j; l < m; ++l) {                        int suml = 0, sump = 0;                        for (int p = i; p <= k; ++p)                            for (int q = j; q <= l; ++q) {                                suml += s[p][q] == 'L';                                sump += s[p][q] == 'P';                            }                        int sum = suml + sump, dif = suml - sump + 1000;                        gao(D[i][dif], sum);                        gao(R[j][dif], sum);                        gao(U[k][dif], sum);                        gao(L[l][dif], sum);                    }        for (int i = 1; i < n; ++i)             for (int j = 0; j < N * N + 1000; ++j)                U[i][j] = max(U[i][j], U[i - 1][j]);        for (int i = n - 2; i >= 0; --i)            for (int j = 0; j < N * N + 1000; ++j)                D[i][j] = max(D[i][j], D[i + 1][j]);        for (int j = 1; j < m; ++j)            for (int k = 0; k < N * N + 1000; ++k)                L[j][k] = max(L[j][k], L[j - 1][k]);        for (int j = m - 2; j >= 0; --j)            for (int k = 0; k < N * N + 1000; ++k)                R[j][k] = max(R[j][k], R[j + 1][k]);        int ans = -1;        for (int i = -n * m; i <= n * m; ++i)            for (int j = -n * m; j <= n * m; ++j) {                for (int h = 0; h < n - 1; ++h)                     if (abs(i + j) <= maxDiff)  ans = max(ans, U[h][i + 1000] + D[h + 1][j + 1000]);                for (int l = 0; l < m - 1; ++l)                    if (abs(i + j) <= maxDiff)  ans = max(ans, L[l][i + 1000] + R[l + 1][j + 1000]);            }        return ans;    }};

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