LOJ#3092. 「BJOI2019」排兵布阵

这题就是个背包啊，感觉是\(nms\)的但是不到0.2s，发生了什么。。

就是设\(f[i]\)为选了\(i\)个人最大的代价，然后有用的人数只有\(s\)种

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 500005
#define ba 47
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int64 f[20005];
int a[105][105],S,N,M;
vector<int> v;
vector<pii > vec;

void upmax(int64 &x,int64 y) {
    x = max(x,y);
}
void Solve() {
    read(S);read(N);read(M);
    for(int i = 1 ;i <= S ; ++i) {
    for(int j = 1 ; j <= N ; ++j) {
        read(a[i][j]);
    }
    }
    for(int j = 1 ; j <= N ; ++j) {
    v.clear();vec.clear();
    for(int i = 1 ; i <= S ; ++i) {
        if(a[i][j] * 2 + 1 <= M) v.pb(a[i][j] * 2 + 1);
    }
    sort(v.begin(),v.end());
    if(v.size()) {
        vec.pb(mp(v[0],1));
        for(int i = 1 ; i < v.size() ; ++i) {
        pii t = vec.back();
        if(v[i] == v[i - 1]) {
            t.se++;vec.pop_back();vec.push_back(t);
        }
        else vec.pb(mp(v[i],t.se + 1));
        }
    }
    for(int i = M ; i >= 0 ; --i) {
        for(auto t : vec) {
        if(i >= t.fi) {
            upmax(f[i],f[i - t.fi] + 1LL * t.se * j);
        }
        else break;
        }
    }
    }
    out(f[M]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}