uva 825 简单DP

题意：

给张图，求左上到右下的可达路径有多少条。

解析：

状态转移方程：

for (int i = 1; i <= w; i++)
{
    for (int j = 1; j <= n; j++)
    {
        if (g[i][j])
        {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
        else
        {
            dp[i][j] = 0;
        }
    }
}

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long

using namespace std;
const int maxn = 1e4 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = 4 * atan(1.0);
const double ee = exp(1.0);

bool g[maxn][maxn];
int dp[maxn][maxn];

int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif // LOCAL
    int ncase;
    scanf("%d", &ncase);
    while (ncase--)
    {
        int w, n;
        scanf("%d%d", &w, &n);
        memset(g, true, sizeof(g));
        memset(dp, 0, sizeof(dp));
        getchar();
        for (int k = 0; k < w; k++)
        {
            int x;
            scanf("%d", &x);
            char t[maxn];
            gets(t);
//            cout << t << endl;
            int len = strlen(t);
            int y = 0;
            for (int j = 0; j <= len; j++)
            {
                if ('0' <= t[j] && t[j] <= '9')
                {
                    y = y * 10 + (t[j] - '0');
                }
                else
                {
//                    cout << y << endl;
                    g[x][y] = false;
//                    cout << y << endl;
                    y = 0;
                }
            }
        }
        dp[1][0] = 1;
        for (int i = 1; i <= w; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (g[i][j])
                {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
                else
                {
                    dp[i][j] = 0;
                }
            }
        }
        printf("%d\n", dp[w][n]);
        if (ncase)
            printf("\n");
    }
    return 0;
}