D. R2D2 and Droid Army(二分加多维线段树查询区间最大值)

简单线段树,直接上代码.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <cstdlib>
#include <algorithm>

#define ls u << 1
#define rs u << 1 | 1
#define lson l, mid, u << 1
#define rson mid + 1, r, u << 1 | 1
#define INF 0x3f3f3f3f
#define MAX 4

using namespace std;
typedef long long ll;
const int M = 1e5 + 100;
const int mod = 2147483647;
int sum[10][M << 2],n,m,k,res[10],ans[10];

void pushup(int u){
    for(int i = 0; i < m; i++)
        sum[i][u] = max(sum[i][ls],sum[i][rs]);
}

void build(int l,int r,int u){
    if(l == r){
        for(int i = 0; i < m; i++)
            scanf("%d",&sum[i][u]);
    }
    else{
        int mid = (l + r) >> 1;
        build(lson);
        build(rson);
        pushup(u);
    }
}

void query(int L,int R,int l,int r,int u){
    if(L <= l && R >= r){
        for(int i = 0; i < m; i++)
            res[i] = max(res[i],sum[i][u]);
    }
    else{
        int mid = (l + r) >> 1;
        if(L <= mid) query(L,R,lson);
        if(R > mid) query(L,R,rson);
    }
}

bool solve(int len){
    for(int l = 1; l <= n - len + 1; l++){
        int r = l + len - 1,sum = 0;
        memset(res,-1,sizeof(res));
        query(l,r,1,n,1);
        for(int i = 0; i < m; i++){
            sum += res[i];
        }
        if(sum <= k){
            for(int i = 0; i < m; i++)
                ans[i] = res[i];
            return true;
        }
    }
    return false;
}

int main(){
    scanf("%d %d %d",&n,&m,&k);
    build(1,n,1);
    int l = 1,r = n;
    while(l <= r){
        int mid = (l + r) >> 1;
        if(solve(mid)) l = mid + 1;
        else r = mid - 1;
    }
    for(int i = 0; i < m; i++){
        if(i) putchar(' ');
        printf("%d",ans[i]);
    }
    puts("");
    return 0;
}