牛客子矩阵最大累加和问题

喜欢ヅ旅行 2023-06-04 14:57 14阅读 0赞

题目链接：[https://www.nowcoder.com/practice/cb82a97dcd0d48a7b1f4ee917e2c0409?tpId=101&tqId=33095&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking][https_www.nowcoder.com_practice_cb82a97dcd0d48a7b1f4ee917e2c0409_tpId_101_tqId_33095_tPage_1_rp_1_ru_ta_programmer-code-interview-guide_qru_ta_programmer-code-interview-guide_question-ranking]

### 题目大意： ###

略。

### 分析： ###

略.

### 代码如下： ###

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1 #include <bits/stdc++.h>
      2 using namespace std;
      3  
      4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      5 #define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
      6 #define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
      7 #define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i)
      8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
      9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
     10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
     11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
     12  
     13 #define pr(x) cout << #x << " = " << x << "  "
     14 #define prln(x) cout << #x << " = " << x << endl
     15  
     16 #define LOWBIT(x) ((x)&(-x))
     17  
     18 #define ALL(x) x.begin(),x.end()
     19 #define INS(x) inserter(x,x.begin())
     20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
     21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
     22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
     23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
     24  
     25 #define ms0(a) memset(a,0,sizeof(a))
     26 #define msI(a) memset(a,0x3f,sizeof(a))
     27 #define msM(a) memset(a,-1,sizeof(a))
     28 
     29 #define MP make_pair
     30 #define PB push_back
     31 #define ft first
     32 #define sd second
     33  
     34 template<typename T1, typename T2>
     35 istream &operator>>(istream &in, pair<T1, T2> &p) {
     36     in >> p.first >> p.second;
     37     return in;
     38 }
     39  
     40 template<typename T>
     41 istream &operator>>(istream &in, vector<T> &v) {
     42     for (auto &x: v)
     43         in >> x;
     44     return in;
     45 }
     46 
     47 template<typename T>
     48 ostream &operator<<(ostream &out, vector<T> &v) {
     49     Rep(i, v.size()) out << v[i] << " \n"[i == v.size() - 1];
     50     return out;
     51 }
     52  
     53 template<typename T1, typename T2>
     54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     55     out << "[" << p.first << ", " << p.second << "]" << "\n";
     56     return out;
     57 }
     58 
     59 inline int gc(){
     60     static const int BUF = 1e7;
     61     static char buf[BUF], *bg = buf + BUF, *ed = bg;
     62     
     63     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
     64     return *bg++;
     65 } 
     66 
     67 inline int ri(){
     68     int x = 0, f = 1, c = gc();
     69     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
     70     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
     71     return x*f;
     72 }
     73 
     74 template<class T>
     75 inline string toString(T x) {
     76     ostringstream sout;
     77     sout << x;
     78     return sout.str();
     79 }
     80 
     81 inline int toInt(string s) {
     82     int v;
     83     istringstream sin(s);
     84     sin >> v;
     85     return v;
     86 }
     87 
     88 //min <= aim <= max
     89 template<typename T>
     90 inline bool BETWEEN(const T aim, const T min, const T max) {
     91     return min <= aim && aim <= max;
     92 }
     93 
     94 typedef unsigned int uI;
     95 typedef long long LL;
     96 typedef unsigned long long uLL;
     97 typedef vector< int > VI;
     98 typedef vector< bool > VB;
     99 typedef vector< char > VC;
    100 typedef vector< double > VD;
    101 typedef vector< string > VS;
    102 typedef vector< LL > VL;
    103 typedef vector< VI > VVI;
    104 typedef vector< VB > VVB;
    105 typedef vector< VS > VVS;
    106 typedef vector< VL > VVL;
    107 typedef vector< VVI > VVVI;
    108 typedef vector< VVL > VVVL;
    109 typedef pair< int, int > PII;
    110 typedef pair< LL, LL > PLL;
    111 typedef pair< int, string > PIS;
    112 typedef pair< string, int > PSI;
    113 typedef pair< string, string > PSS;
    114 typedef pair< double, double > PDD;
    115 typedef vector< PII > VPII;
    116 typedef vector< PLL > VPLL;
    117 typedef vector< VPII > VVPII;
    118 typedef vector< VPLL > VVPLL;
    119 typedef vector< VS > VVS;
    120 typedef map< int, int > MII;
    121 typedef unordered_map< int, int > uMII;
    122 typedef map< LL, LL > MLL;
    123 typedef map< string, int > MSI;
    124 typedef map< int, string > MIS;
    125 typedef multiset< int > mSI;
    126 typedef multiset< char > mSC;
    127 typedef set< int > SI;
    128 typedef stack< int > SKI;
    129 typedef stack< char > SKC;
    130 typedef deque< int > DQI;
    131 typedef queue< int > QI;
    132 typedef priority_queue< int > PQIMax;
    133 typedef priority_queue< int, VI, greater< int > > PQIMin;
    134 const double EPS = 1e-8;
    135 const LL inf = 0x7fffffff;
    136 const LL infLL = 0x7fffffffffffffffLL;
    137 const LL mod = 1e9 + 7;
    138 const int maxN = 2e2 + 7;
    139 const LL ONE = 1;
    140 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
    141 const LL oddBits = 0x5555555555555555;
    142 
    143 int N, M;
    144 int ans; 
    145 int arr[maxN][maxN];
    146 int preSum[maxN][maxN];
    147 
    148 // 求对角点为(x1, y1)和(x2, y2)的矩阵累加和 
    149 inline int getSum(int x1, int y1, int x2, int y2) {
    150     return preSum[x2][y2] - preSum[x1 - 1][y2] - preSum[x2][y1 - 1] + preSum[x1 - 1][y1 - 1];
    151 }
    152 
    153 int main() {
    154     //freopen("MyOutput.txt","w",stdout);
    155     //freopen("input.txt","r",stdin);
    156     //INIT();
    157     scanf("%d%d", &N, &M);
    158     
    159     // 由于复杂度是O(M*N^2),N应该选择min(N, M) 
    160     if(N <= M) {
    161         For(i, 1, N) {
    162             For(j, 1, M) {
    163                 scanf("%d", &arr[i][j]);
    164             }
    165         }
    166     }
    167     else {
    168         For(i, 1, N) {
    169             For(j, 1, M) {
    170                 scanf("%d", &arr[j][i]);
    171             }
    172         }
    173         swap(N, M);
    174     }
    175     ans = arr[1][1];
    176     
    177     // 计算二维前缀和 
    178     For(i, 1, N) {
    179         For(j, 1, M) {
    180             preSum[i][j] = arr[i][j] + preSum[i - 1][j] + preSum[i][j - 1] - preSum[i - 1][j - 1];
    181         }
    182     }
    183     
    184     For(i, 1, N) { // 枚举矩阵的宽 
    185         For(j, 1, N - i + 1) { // 枚举矩阵顶部所在行 
    186             // 接下来就变成子数组最大累加和问题 
    187             int ret = 0;
    188             
    189             For(k, 1, M) {
    190                 ret += getSum(j, k, j + i - 1, k);
    191                 ans = max(ans, ret);
    192                 if(ret <= 0) ret = 0;
    193             } 
    194         }
    195     } 
    196     
    197     printf("%d\n", ans);
    198     return 0;
    199 }

转载于:https://www.cnblogs.com/zaq19970105/p/11528798.html

[https_www.nowcoder.com_practice_cb82a97dcd0d48a7b1f4ee917e2c0409_tpId_101_tqId_33095_tPage_1_rp_1_ru_ta_programmer-code-interview-guide_qru_ta_programmer-code-interview-guide_question-ranking]: https://www.nowcoder.com/practice/cb82a97dcd0d48a7b1f4ee917e2c0409?tpId=101&tqId=33095&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking
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