OJ-Problems-Source/HDOJ/4331_autoAC.cpp

126 lines
3.1 KiB
C++
Raw Normal View History

#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#define MAXN 1005
using namespace std;
int N, G[MAXN][MAXN], c[MAXN], idx;
int dgn[MAXN];
int l[MAXN][MAXN], r[MAXN][MAXN], u[MAXN][MAXN], d[MAXN][MAXN];
struct Node
{
int x, sign, key;
bool operator < (Node temp) const
{
if (x != temp.x) {
return x < temp.x;
}
else {
return sign < temp.sign;
}
}
}p[MAXN<<1];
int lowbit(int x)
{
return x & -x;
}
void addpoint(int a, int b)
{
++idx;
p[idx].x = a, p[idx].sign = 0;
p[idx].key = a;
++idx;
p[idx].x = b, p[idx].sign = 1;
p[idx].key = a;
}
void add(int x)
{
for (int i = x; i <= N; i += lowbit(i)) {
c[i] += 1;
}
}
int sum(int x)
{
int ret = 0;
for (int i = x; i > 0; i -= lowbit(i)) {
ret += c[i];
}
return ret;
}
int solve()
{
int ret = 0;
for (int i = N; i >= 1; --i) {
idx = -1;
for (int j = 1; j <= (N-i+1); ++j) {
if (G[i+j-1][j]) {
addpoint(j, j+min(r[i+j-1][j], d[i+j-1][j])-1);
dgn[j] = j - min(u[i+j-1][j], l[i+j-1][j])+1;
}
}
if (idx == -1) { continue; }
sort(p, p+idx+1);
memset(c, 0, sizeof (c));
for (int k = 0; k <= idx; ++k) {
if (!p[k].sign) {
ret -= sum(p[k].key);
add(dgn[p[k].x]);
}
else {
ret += sum(p[k].key);
}
}
}
for (int j = 2; j <= N; ++j) {
idx = -1;
for (int i = 1; i <= (N-j+1); ++i) {
if (G[i][j+i-1]) {
addpoint(i, i+min(r[i][j+i-1], d[i][j+i-1])-1);
dgn[i] = i - min(u[i][j+i-1], l[i][j+i-1])+1;
}
}
if (idx == -1) { continue; }
sort(p, p+idx+1);
memset(c, 0, sizeof (c));
for (int k = 0; k <= idx; ++k) {
if (!p[k].sign) {
ret -= sum(p[k].key);
add(dgn[p[k].x]);
}
else {
ret += sum(p[k].key);
}
}
}
return ret;
}
int main()
{
int T, ca = 0;
scanf("%d", &T);
while (T--) {
scanf("%d", &N);
memset(d, 0, sizeof (d));
memset(r, 0, sizeof (r));
for (int i = 1; i<= N; ++i) {
for (int j = 1; j <= N; ++j) {
scanf("%d", &G[i][j]);
if (G[i][j]) {
u[i][j] = G[i-1][j] ? u[i-1][j] + 1 : 1;
l[i][j] = G[i][j-1] ? l[i][j-1] + 1 : 1;
}
}
}
for (int i = N; i >= 1; --i) {
for (int j = N; j >= 1; --j) {
if (G[i][j]) {
d[i][j] = G[i+1][j] ? d[i+1][j] + 1 : 1;
r[i][j] = G[i][j+1] ? r[i][j+1] + 1 : 1;
}
}
}
printf("Case %d: %d\n", ++ca, solve());
}
return 0;
}