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ID提交记录ID题目HackerOwner结果提交时间测评时间
#145#197806#3433. Crisscrosssnow_traceZeardoeFailed.2024-02-18 12:59:352024-02-18 12:59:37

详细

Extra Test:

Invalid Input

input:

114514

output:


result:

FAIL Integer 114514 violates the range [1, 1000] (stdin)

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#197806#3433. CrisscrossZeardoe100108ms2268kbC++112.0kb2023-11-14 16:25:182023-11-14 17:29:24

answer

/*
[templates]: 
duipai
spjdp
compre
addhis
floor_sum
treedfs
matrix
network_flow
polynomial
lca
bitset
valuesgt
fenwick
erbitree
*/
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx")
#include<bits/stdc++.h>
using namespace std;
#define int long long
//use ll instead of int.
#define f(i, a, b) for(int i = (a); i <= (b); i++)
#define cl(i, n) i.clear(),i.resize(n);
#define endl '\n'
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int inf = 1e18;
//#define cerr if(false)cerr
//#define freopen if(false)freopen
mt19937 rng(time(0)); 
int rnd(int l, int r) {return rng() % (r-l+1) + l; }
#define watch(x) cerr  << (#x) << ' '<<'i'<<'s'<<' ' << x << endl
void pofe(int number, int bitnum) {
    string s; f(i, 0, bitnum) {s += char(number & 1) + '0'; number >>= 1; } 
    reverse(s.begin(), s.end()); cerr << s << endl; 
    return;
}
template <typename TYP> void cmax(TYP &x, TYP y) {if(x < y) x = y;}
template <typename TYP> void cmin(TYP &x, TYP y) {if(x > y) x = y;}
//调不出来给我对拍!
//use std::array.
const int N = 1000; 
char a[N + 10][N + 10]; 
int cntt[N + 10], cnttt[N + 10];
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    //freopen();
    //freopen();
    //time_t start = clock();
    //think twice,code once.
    //think once,debug forever.
    int n, m; cin >> n >> m; 
    f(i, 1, n) f(j, 1, m) cin >> a[i][j]; 
    int ans = 0; 
    f(i, 1, n) {
        int cnt = 0; f(j, 1, m) cnt += a[i][j] == '0';
        cntt[i] = cnt; 
    }
    f(i, 1, m) {
        int cnt = 0; f(j, 1, n) cnt += a[j][i] == '0';
        cnttt[i] = cnt; 
    }
    f(i, 1, n) ans += cntt[i] > 0;
    f(i, 1, m) ans += cnttt[i] > 0;
    f(i, 1, n) f(j, 1, m) ans -= cntt[i] == 1 && cnttt[j] == 1 && a[i][j] == '0'; 
    cout << (int)(ans + 1)  << endl; 
    //time_t finish = clock();
    //cout << "time used:" << (finish-start) * 1.0 / CLOCKS_PER_SEC <<"s"<< endl;
    return 0;
}