Codeforces 677E Vanya and Balloons(暴力+转换) category:

题意：

给定一个矩阵，让你其中有1,2,3,0，让你在里面寻找一个十字，是的这个十字所包含的数字的乘积最大。

解法：

这个完全可以预处理处八个方向的前缀，然后暴力枚举中心即可，要注意的是，十字的四条边长是一样的，另外因为数字太大，所以用log把乘变成加，就可以比较大小了，不然直接模是无法比大小的。代码略长。。。

//
//  Created by  CQU_CST_WuErli
//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.
//
//#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define CLR(x) memset(x,0,sizeof(x))
#define OFF(x) memset(x,-1,sizeof(x))
#define MEM(x,a) memset((x),(a),sizeof(x))
#define BUG cout << "I am here" << endl
#define lookln(x) cout << #x << "=" << x << endl
#define SI(a) scanf("%d", &a)
#define SII(a,b) scanf("%d%d", &a, &b)
#define SIII(a,b,c) scanf("%d%d%d", &a, &b, &c)
const int INF_INT=0x3f3f3f3f;
const long long INF_LL=0x7f7f7f7f;
const int MOD=1e9+7;
const double eps=1e-10;
const double pi=acos(-1);
typedef long long  ll;
using namespace std;

const int N = 1010;
const double lg2 = log(2);
const double lg3 = log(3);
int n;
char a[N][N];

double logs[N][N];
double L[N][N], R[N][N], U[N][N], D[N][N];
double LU[N][N], RU[N][N], LD[N][N], RD[N][N];

int preL[N][N], preR[N][N], preU[N][N], preD[N][N];
int preLU[N][N], preRU[N][N], preLD[N][N], preRD[N][N];

void gao() {
    CLR(L);
    CLR(preL);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) if (a[i][j] != '0') {
            if (j == 1 || a[i][j - 1] == '0') {
                L[i][j] = logs[i][j];
                preL[i][j] = 1;
            }
            else {
                L[i][j] = L[i][j - 1] + logs[i][j];
                preL[i][j] = preL[i][j - 1] + 1;
            }
        }
    }
    CLR(R);
    CLR(preR);
    for (int i = 1; i <= n; i++) {
        for (int j = n; j >= 1; j--) if (a[i][j] != '0') {
            if (j == n || a[i][j + 1] == '0') {
                R[i][j] = logs[i][j];
                preR[i][j] = 1;
            }
            else {
                R[i][j] = R[i][j + 1] + logs[i][j];
                preR[i][j] = preR[i][j + 1] + 1;
            }
        }
    }
    CLR(U);
    CLR(preU);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) if (a[i][j] != '0') {
            if (i == 1 || a[i - 1][j] == '0')  {
                U[i][j] = logs[i][j];
                preU[i][j] = 1;
            }
            else {
                U[i][j] = U[i - 1][j] + logs[i][j];
                preU[i][j] = preU[i - 1][j] + 1;
            }
        }
    }
    CLR(D);
    CLR(preD);
    for (int i = n; i >= 1; i--) {
        for (int j = 1; j <= n; j++) if (a[i][j] != '0') {
            if (i == n || a[i + 1][j] == '0') {
                D[i][j] = logs[i][j];
                preD[i][j] = 1;
            }
            else {
                D[i][j] = D[i + 1][j] + logs[i][j];
                preD[i][j] = preD[i + 1][j] + 1;
            }
        }
    }
    CLR(LU);
    CLR(preLU);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) if (a[i][j] != '0') {
            if (i == 1 || j == 1 || a[i - 1][j - 1] == '0') {
                LU[i][j] = logs[i][j];
                preLU[i][j] = 1;
            }
            else {
                LU[i][j] = LU[i - 1][j - 1] + logs[i][j];
                preLU[i][j] = preLU[i - 1][j - 1] + 1;
            }
        }
    }
    CLR(RU);
    CLR(preRU);
    for (int i = 1; i <= n; i++) {
        for (int j = n; j >= 1; j--) if (a[i][j] != '0') {
            if (i == 1 || j == n || a[i - 1][j + 1] == '0') {
                RU[i][j] = logs[i][j];
                preRU[i][j] = 1;
            }
            else {
                RU[i][j] = RU[i - 1][j + 1] + logs[i][j];
                preRU[i][j] = preRU[i - 1][j + 1] + 1;
            }
        }
    }
    CLR(LD);
    CLR(preLD);
    for (int i = n; i >= 1; i--) {
        for (int j = 1; j <= n; j++) if (a[i][j] != '0') {
            if (i == n || j == 1 || a[i + 1][j - 1] == '0') {
                LD[i][j] = logs[i][j];
                preLD[i][j] = 1;
            }
            else {
                LD[i][j] = LD[i + 1][j - 1] + logs[i][j];
                preLD[i][j] = preLD[i + 1][j - 1] + 1;
            }
        }
    }
    CLR(RD);
    CLR(preRD);
    for (int i = n; i >= 1; i--) {
        for (int j = n; j >= 1; j--) if (a[i][j] != '0') {
            if (i == n || j == n || a[i + 1][j + 1] == '0') {
                RD[i][j] = logs[i][j];
                preRD[i][j] = 1;
            }
            else {
                RD[i][j] = RD[i + 1][j + 1] + logs[i][j];
                preRD[i][j] = preRD[i + 1][j + 1] + 1;
            }
        }
    }
}

int main(int argc, char const *argv[]) {
#ifdef LOCAL
    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);
    // freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);
#endif
    while (SI(n) == 1) {
        CLR(logs);
        for (int i = 1; i <= n; i++)
            scanf("%s", a[i] + 1);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (a[i][j] == '0') logs[i][j] = -1;
                else if (a[i][j] == '1') logs[i][j] = 0;
                else if (a[i][j] == '2') logs[i][j] = lg2;
                else if (a[i][j] == '3') logs[i][j] = lg3;
            }
        }
        gao();
        double ans = -1;
        pair pos;
        int flag = 0;
        int Len;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) if (a[i][j] != '0') {
                double tmp = 0;
                int len = min(min(preL[i][j], preR[i][j]), min(preU[i][j], preD[i][j]));
                tmp += L[i][j] - L[i][j - len];
                tmp += R[i][j] - R[i][j + len];
                tmp += U[i][j] - U[i - len][j];
                tmp += D[i][j] - D[i + len][j];
                tmp -= 3 * logs[i][j];
                if (tmp > ans) {
                    ans = tmp;
                    pos = make_pair(i, j);
                    flag = 1;
                    Len = len;
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) if (a[i][j] != '0') {
                double tmp = 0;
                int len = min(min(preLD[i][j], preRD[i][j]), min(preLU[i][j], preRU[i][j]));
                tmp += LU[i][j] - LU[i - len][j - len];
                tmp += RU[i][j] - RU[i - len][j + len];
                tmp += LD[i][j] - LD[i + len][j - len];
                tmp += RD[i][j] - RD[i + len][j + len];
                tmp -= 3 * logs[i][j];
                if (tmp > ans) {
                    ans = tmp;
                    pos = make_pair(i, j);
                    flag = 2;
                    Len = len;
                }
            }
        }
        if (ans == -1) {
            cout << 0 << endl;
            continue;
        }
        int x = pos.first, y = pos.second;
        ll tmp = 1;
        if (flag == 1) {
            for (int i = 1; i < Len; i++) {
                tmp = tmp * (a[x][y - i] - '0') % MOD;
                tmp = tmp * (a[x][y + i] - '0') % MOD;
                tmp = tmp * (a[x - i][y] - '0') % MOD;
                tmp = tmp * (a[x + i][y] - '0') % MOD;
            }
            tmp = tmp * (a[x][y] - '0') % MOD;
        }
        else {
            for (int i = 1; i < Len; i++) {
                tmp = tmp * (a[x - i][y - i] - '0') % MOD;
                tmp = tmp * (a[x - i][y + i] - '0') % MOD;
                tmp = tmp * (a[x + i][y - i] - '0') % MOD;
                tmp = tmp * (a[x + i][y + i] - '0') % MOD;
            }
            tmp = tmp * (a[x][y] - '0') % MOD;
        }
        cout << tmp << endl;
    }
    return 0;
}