D. R2D2 and Droid Army
time limit per test 2 seconds
memory limit per test
An army of n droids is lined up in one row. Each droid is described bym integers a1,a2,…,am, whereai is the number of details of thei-th type in this droid’s mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He hasm weapons, the i-th weapon can affect all the droids in the army by destroying one detail of thei-th type (if the droid doesn’t have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at mostk shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n,m,k (1≤n≤105,1≤m≤5, 0≤k≤109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line containsm integers a1,a2,…,am (0≤ai≤108), where ai is the number of details of thei-th type for the respective robot.
Output
Print m space-separated integers, where thei-th number is the number of shots from the weapon of thei-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Sample test(s)
Input
5 2 44 01 22 10 21 3
Output
2 2
Input
3 2 41 21 32 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
AC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 4using namespace std;typedef long long ll;const int M = 1e5 + 100;const int mod = 2147483647;multiset<int>mp[10];int a[M][10],res[10];int main(){int n,m,k,len = 0;cin>>n>>m>>k;for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)scanf("%d",&a[i][j]);for(int l = 0,r = 0; r < n; r++){for(int i = 0; i < m; i++){mp[i].insert(a[r][i]);}while(l <= r){int sum = 0;for(int i = 0; i < m; i++)sum += *mp[i].rbegin();if(sum <= k) break;for(int i = 0; i < m; i++)mp[i].erase(mp[i].find(a[l][i]));l++;}if(len < r – l + 1){len = r – l + 1;for(int i = 0; i < m; i++)res[i] = *mp[i].rbegin();}}for(int i = 0; i < m; i++){if(i) putchar(' ');printf("%d",res[i]);}return 0;}
，也只有懂的接受自己的失败，才能更好的去发挥自身优势，也才能够更好的去实现自我;
