uva 825 - Walking on the Safe Side(dp)

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题目链接：825 - Walking on the Safe Side

题目大意：给出n,m,现在给出n行数据，每行有k(k为不定值）个数字，第一个数字代表行数，后面k - 1个数代表当前行的这个位置不可走，问有多少路径可以从（1，1）到（n,m）,只能向下或向右。

解题思路：dp[i][j] = dip[i - 1][j] + dp[i][j - 1], 很简单的dp问题。

#include <stdio.h>#include <string.h>const int N = 1005;int n, m, dp[N][N], g[N][N];void handle(int k, char str[]) {int len = strlen(str), num = 0;for (int i = 0; i <= len; i++) {if (str[i] >= '0' && str[i] <= '9')num = num * 10 + str[i] - '0';else {g[k][num] = 1;num = 0;}}}void read() {int r;char str[N];memset(dp, 0, sizeof(dp));memset(g, 0, sizeof(g));scanf("%d%d", &n, &m);for (int i = 0; i < n; i++) {scanf("%d", &r);gets(str);handle(r, str);}}int solve () {dp[0][1] = 1;for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {if (g[i][j])continue;dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}return dp[n][m];}int main() {int cas;scanf("%d", &cas);while (cas--) {read();printf("%d\n", solve());if (cas) printf("\n");}return 0;}