本文共 1098 字，大约阅读时间需要 3 分钟。
考虑k比较小，想到容斥原理，枚举，容斥求和。
//http://www.cnblogs.com/IMGavin/#include #include #include #include #include #include #include #include #include #include #include using namespace std;typedef long long LL;#define gets(A) fgets(A, 1e8, stdin)const int INF = 0x3F3F3F3F, N = 1008, MOD = 1003;const double EPS = 1e-6;int x[N], y[N];int tot;int r[N], c[N], n, m, k;LL ans;void dfs(int d, int cnt, int x1, int y1){ if(d == tot){ if(cnt & 1){ ans -= x1 * y1; }else{ ans += x1 * y1; } return ; } dfs(d + 1, cnt, x1, y1); dfs(d + 1, cnt + 1, min(x1, x[d]), min(y1, y[d]));}int main(){ while(cin >> n >> m >> k){ for(int i = 0; i < k; i++){ scanf("%d %d", &r[i], &c[i]); } ans = 0; for(int i = 1; i <= n; i++){ for(int j = 1; j <= m; j++){ tot = 0; bool ok = 1; for(int l = 0; l < k; l++){ if(i == r[l] && j == c[l]){ ok = 0; break; } if(i >= r[l] && j >= c[l]){ x[tot] = r[l]; y[tot] = c[l]; tot++; } } if(ok){ dfs(0, 0, i, j); } } } cout< <
转载于:https://www.cnblogs.com/IMGavin/p/6505755.html