dp学习之AvoidRoads解法二（算法优化）

#include <iostream>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <sstream>

using namespace std;
int dx[] = {1, 0};
int dy[] = {0, 1};
class AvoidRoads
{
public:
long long numWays(int width, int height, vector<string> &bad);
};

int main(void)
{
AvoidRoads l;
vector<string> str;
#if 0
str.push_back("0 3 0 2");
str.push_back("1 2 1 3");
str.push_back("2 2 2 3");
str.push_back("3 2 3 3");
str.push_back("4 2 4 3");
str.push_back("5 2 5 3");
str.push_back("6 2 6 3");
str.push_back("7 2 7 3");
str.push_back("8 2 8 3");
str.push_back("9 2 9 3");
#endif
str.push_back("0 0 0 1");
str.push_back("6 6 5 6");
cout << l.numWays(6, 6, str) << endl;
return 0;
}
long long AvoidRoads::numWays(int width, int height, vector<string> &bad)
{
long long D[width + 1][height + 1];
char is_bad1[128], is_bad2[128];
for(int i = 0; i <= width; i++)
{
for(int j = 0; j <= height; j++)
{
D[i][j] = 0;
}
}
D[0][0] = 1;
for(int i = 0; i <= width; i++)
{
for(int j = 0; j <= height; j++)
{
for(int k = 0; k < 2; k++)
{
int is_bad = 0;
int tx = i + dx[k];
int ty = j + dy[k];
#if 1        /*If you know why the next two line is needful, you are so clever*/
if(tx > width || ty > height)
continue;
#endif
snprintf(is_bad1, sizeof(is_bad1), "%d %d %d %d", i, j, tx, ty);
snprintf(is_bad2, sizeof(is_bad2), "%d %d %d %d", tx, ty, i, j);
for(int p = 0; p < bad.size(); p++)
{
if(bad[p] == string(is_bad1) || bad[p] == string(is_bad2))
{
is_bad = 1;
break;
}
}
if(is_bad != 1)
{
D[tx][ty] += D[i][j];
}
}
}
}
return D[width][height];
}