#include <cstdio>
#include <cstring>
using namespace std;
const int N = 100;
short g[N][N];
bool check(int x1, int y1, int x2, int y2);
int main()
{
int n, m;
char ch;
int x1, y1, x2, y2;
int cnt;
#ifndef ONLINE_JUDGE
freopen("uva_in.txt", "r", stdin);
#endif
while (scanf("%d", &n) && n) {
scanf("%d", &m);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; ) {
ch = getchar();
if (ch == '0' || ch == '1') {
g[i][j++] = ch - '0';
}
}
}
cnt = 0;
for (x1 = 0; x1 < n; x1++) {
for ( y1 = 0; y1 < m; y1++) {
for (x2 = x1; x2 < n; x2++) {
for (y2 = y1; y2 < m; y2++) {
if (check(x1, y1, x2, y2)) cnt++;
}
}
}
}
printf("%d\n", cnt);
}
return 0;
}
bool check(int x1, int y1, int x2, int y2)
{
for (int i = x1; i <= x2; i++) {
for (int j = y1; j <= y2; j++) {
if (g[i][j] != 1) return false;
}
}
return true;
}
上面用的是枚举法,效率不高,改成组合的计数方法,代码如下
#include <cstdio>
#include <algorithm>
#include <climits>
using namespace std;
#pragma warning(disable:4996)
const int N = 110;
int dp[N][N];
int row, col;
char buf[N];
bool input()
{
scanf("%d", &row);
if (row == 0) return false;
scanf("%d", &col);
for (int i = 1; i <= row; i++) {
scanf("%s", buf);
for (int j = 1; j <= col; j++) {
if (buf[j - 1] == '1') dp[i][j] = 1;
else dp[i][j] = 0;
}
}
return true;
}
void solve()
{
for (int i = 2; i <= row; i++) {
for (int j = 1; j <= col; j++) {
if (dp[i][j]) dp[i][j] += dp[i - 1][j];
}
}
int ans = 0;
for (int i = 1; i <= row; i++) {
for(int j = 1; j <= col; j++) {
int tmp = INT_MAX;
for (int k = j; k <= col && dp[i][k]; k++) {
tmp = min(tmp, dp[i][k]);
ans += tmp;
}
}
}
printf("%d\n", ans);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("f:\\OJ\\uva_in.txt", "r", stdin);
#endif
while (input()) {
solve();
}
return 0;
}