牛客 子矩阵最大累加和问题

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

题目大意:

  略。

分析:

   略.

代码如下:

  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] << " 
"[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 << "]" << "
";
 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
", ans);
198     return 0;
199 }
View Code
原文地址:https://www.cnblogs.com/zaq19970105/p/11528798.html