• 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

      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 < (n); ++i)
      6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
      7 #define rFor(i,t,s) for (int i = (t); i >= (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()); // ?? x ?????? 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,inf,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()];
     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 long long LL;
     95 typedef unsigned long long uLL;
     96 typedef pair< double, double > PDD;
     97 typedef pair< int, int > PII;
     98 typedef pair< int, PII > PIPII;
     99 typedef pair< string, int > PSI;
    100 typedef pair< int, PSI > PIPSI;
    101 typedef set< int > SI;
    102 typedef set< PII > SPII;
    103 typedef vector< int > VI;
    104 typedef vector< double > VD;
    105 typedef vector< VI > VVI;
    106 typedef vector< SI > VSI;
    107 typedef vector< PII > VPII;
    108 typedef map< int, int > MII;
    109 typedef map< int, string > MIS;
    110 typedef map< int, PII > MIPII;
    111 typedef map< PII, int > MPIII;
    112 typedef map< string, int > MSI;
    113 typedef map< string, string > MSS;
    114 typedef map< PII, string > MPIIS;
    115 typedef map< PII, PII > MPIIPII;
    116 typedef multimap< int, int > MMII;
    117 typedef multimap< string, int > MMSI;
    118 //typedef unordered_map< int, int > uMII;
    119 typedef pair< LL, LL > PLL;
    120 typedef vector< LL > VL;
    121 typedef vector< VL > VVL;
    122 typedef priority_queue< int > PQIMax;
    123 typedef priority_queue< int, VI, greater< int > > PQIMin;
    124 const double EPS = 1e-8;
    125 const LL inf = 0x7fffffff;
    126 const LL infLL = 0x7fffffffffffffffLL;
    127 const LL mod = 1e9 + 7;
    128 const int maxN = 1e3 + 7;
    129 const LL ONE = 1;
    130 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
    131 const LL oddBits = 0x5555555555555555;
    132 
    133 int T, N, M, c[maxN][maxN], d[maxN];
    134 LL ans;
    135 // dp[i][j]表示 i 号武器,从等级 j - 1开始能获得的最大收益 
    136 // dp[i][j] = max(dp[i][j + 1] + c[i][j - 1], 0)
    137 LL dp[maxN][maxN];
    138 
    139 // preSum[i] 表示所有武器升到 i 级所能获得的基础分 
    140 LL preSum[maxN];
    141 
    142 int main(){
    143     //freopen("MyOutput.txt","w",stdout);
    144     //freopen("input.txt","r",stdin);
    145     INIT();
    146     cin >> T;
    147     For(cases, 1, T) {
    148         ans = 0;
    149         ms0(preSum);
    150         
    151         cin >> N >> M;
    152         For(i, 1, N) {
    153             For(j, 1, M) {
    154                 cin >> c[i][j];
    155                 c[i][j] = -c[i][j];
    156                 preSum[j] += c[i][j];
    157             }
    158             dp[i][M] = max(0, c[i][M]);
    159         }
    160         
    161         For(j, 1, M) {
    162             cin >> d[j];
    163             preSum[j] += d[j] + preSum[j - 1];
    164         }
    165         
    166         For(i, 1, N) {
    167             rFor(j, M - 1, 1) {
    168                 dp[i][j] = max((LL)0, c[i][j] + dp[i][j + 1]);
    169             }
    170         }
    171         
    172         // 枚举所有武器最低等级 
    173         For(j, 0, M) {
    174             LL sum = preSum[j], minS = infLL;
    175             
    176             For(i, 1, N) {
    177                 sum += dp[i][j + 1];
    178                 minS = min(minS, dp[i][j + 1]);
    179             } 
    180             sum -= minS; // 让收益最低的武器成为短板 
    181             
    182             ans = max(ans, sum);
    183         }
    184         
    185         cout << "Case #" << cases << ": " << ans << endl;
    186     }
    187     return 0;
    188 }
    View Code
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  • 原文地址:https://www.cnblogs.com/zaq19970105/p/11299263.html
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