UVaLive 6854 City (暴力)

Posted 2020-08-09 dwtfukgv

题意：给定一个 n*m 的矩阵，表示有多少条道路与它相连，其中有一个-1，表示未知，道路只能横着和竖着，求-1处的值。

析：根据题意可知，一个点，与其他周围的四个点都可能相连的，也就是说肯定有共用道路的，所以，我们只要算四个点的数就好，然后依次推一下，就得到答案，

也就是说这里面的点可以分为两部分，一部分加起来减去另一部分，就是答案。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d %d", &n, &m);
        int ans = 0, val;
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j){
                scanf("%d", &val);
                if(val <= 0)  continue;
                if(i+j & 1)  ans += val;
                else ans -= val;
            }

        printf("%d\n", abs(ans));
    }
    return 0;
}