2019牛客暑期多校训练营（第六场）- Upgrading Technology

模拟 + 二维st表

枚举每一次所有技能全部到达的level，然后分别对每个技能花费为负的情况往上加，前提是至少有一个不能加，因为可能下一次的全部技能到达的level拿到的收益是个负数。

每次枚举都不拿多的d才可以可以遍历所有情况。

分别枚举每个技能的最小cost的时候，可以用st表来维护前缀和。

#include <bits/stdc++.h>
#define INF 23333333333333333
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 2000;
int n, m, d[N], _, c[N][N], cnt[N], cs, lb[N];
LL pre[N][N], st[N][N][11], b[N];

LL query(int k, int l, int r){
    int t = lb[r - l + 1];
    return min(st[k][l][t], st[k][r - (1 << t) + 1][t]);
}

int main(){

    lb[0] = -1; for(int i = 1; i < N; i ++) lb[i] = lb[i>>1] + 1;
    for(scanf("%d", &_); _; _ --){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i ++){
            for(int j = 1; j <= m; j ++) scanf("%d", &c[i][j]);
        }
        for(int i = 1; i <= m; i ++) scanf("%d", &d[i]);
        for(int i = 1; i <= n; i ++){
            for(int j = 1; j <= m; j ++) pre[i][j] = pre[i][j - 1] + c[i][j];
        }
        for(int i = 1; i <= m; i ++) b[i] = b[i - 1] + d[i];
        for(int i = 1; i <= n; i ++){
            for(int j = 1; j <= m; j ++) st[i][j][0] = pre[i][j];
        }
        for(int k = 1; k <= n; k ++){
            for(int j = 1; j < 11; j ++){
                for(int i = 1; (1 << j) + i - 1 <= m; i ++)
                    st[k][i][j] = min(st[k][i][j - 1], st[k][i + (1 << (j - 1))][j - 1]);
            }
        }
        vector<LL> ans;
        for(int k = 0; k <= m; k ++){
            LL val = b[k];
            for(int i = 1; i <= n; i ++) val -= pre[i][k];
            if(k != m){
                LL sub = -INF;
                bool flag = false;
                for(int i = 1; i <= n; i ++){
                    // k + 1...m minimum cost
                    LL x = query(i, k + 1, m) - pre[i][k];
                    sub = max(sub, x);
                    if(x > 0) flag = true;
                    if(x <= 0) val -= x;
                }
                if(!flag) val += sub;
            }
            ans.pb(val);
        }
        printf("Case #%d: %lld
", ++cs, *max_element(ans.begin(), ans.end()));
    }
    return 0;
}