2019 牛客多校第六场 J Upgrading Technology

题目链接：https://ac.nowcoder.com/acm/contest/886/J

题目大意

　　略。

分析

　　见代码。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())
 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // ?? x ?????? c
 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T>
 ostream &operator<<(ostream &out, vector<T> &v) {
     Rep(i, v.size()) out << v[i] << " \n"[i == v.size()];
     return out;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 template<class T>
 inline string toString(T x) {
     ostringstream sout;
     sout << x;
     return sout.str();
 }

 inline int toInt(string s) {
     int v;
     istringstream sin(s);
     sin >> v;
     return v;
 }

 //min <= aim <= max
 template<typename T>
 inline bool BETWEEN(const T aim, const T min, const T max) {
     return min <= aim && aim <= max;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< int, PII > PIPII;
 typedef pair< string, int > PSI;
 typedef pair< int, PSI > PIPSI;
 typedef set< int > SI;
 typedef set< PII > SPII;
 typedef vector< int > VI;
 typedef vector< double > VD;
 typedef vector< VI > VVI;
 typedef vector< SI > VSI;
 typedef vector< PII > VPII;
 typedef map< int, int > MII;
 typedef map< int, string > MIS;
 typedef map< int, PII > MIPII;
 typedef map< PII, int > MPIII;
 typedef map< string, int > MSI;
 typedef map< string, string > MSS;
 typedef map< PII, string > MPIIS;
 typedef map< PII, PII > MPIIPII;
 typedef multimap< int, int > MMII;
 typedef multimap< string, int > MMSI;
 //typedef unordered_map< int, int > uMII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 typedef priority_queue< int > PQIMax;
 typedef priority_queue< int, VI, greater< int > > PQIMin;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 1e3 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 int T, N, M, c[maxN][maxN], d[maxN];
 LL ans;
 // dp[i][j]表示 i 号武器，从等级 j - 1开始能获得的最大收益
 // dp[i][j] = max(dp[i][j + 1] + c[i][j - 1], 0)
 LL dp[maxN][maxN];

 // preSum[i] 表示所有武器升到 i 级所能获得的基础分
 LL preSum[maxN];

 int main(){
     //freopen("MyOutput.txt","w",stdout);
     //freopen("input.txt","r",stdin);
     INIT();
     cin >> T;
     For(cases, , T) {
         ans = ;
         ms0(preSum);

         cin >> N >> M;
         For(i, , N) {
             For(j, , M) {
                 cin >> c[i][j];
                 c[i][j] = -c[i][j];
                 preSum[j] += c[i][j];
             }
             dp[i][M] = max(, c[i][M]);
         }

         For(j, , M) {
             cin >> d[j];
             preSum[j] += d[j] + preSum[j - ];
         }

         For(i, , N) {
             rFor(j, M - , ) {
                 dp[i][j] = max((LL), c[i][j] + dp[i][j + ]);
             }
         }

         // 枚举所有武器最低等级
         For(j, , M) {
             LL sum = preSum[j], minS = infLL;

             For(i, , N) {
                 sum += dp[i][j + ];
                 minS = min(minS, dp[i][j + ]);
             }
             sum -= minS; // 让收益最低的武器成为短板 

             ans = max(ans, sum);
         }

         cout << "Case #" << cases << ": " << ans << endl;
     }
     return ;
 }