UVA - 825 Walking on the Safe Side

来源：互联网发布：价格表制作软件编辑：程序博客网时间：2024/05/16 08:12

题意：求从坐标（1，1）到（n,m）的路径数，有些地方是不能走的，输入有点麻烦而已，只能走右或下，所以动态转移方程就很简单了dp[i][j] = dp[i-1][j] + dp[i][j-1]

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 1010;char str[MAXN];int n,m,G[MAXN][MAXN],dp[MAXN][MAXN];int main(){    int t;    scanf("%d",&t);    while (t--){        scanf("%d%d%*c",&n,&m);        memset(G,1,sizeof(G));        memset(dp,0,sizeof(dp));        for (int i = 1; i <= n; i++){            int tmp;            scanf("%d",&tmp);            gets(str);            int len = strlen(str);            int cur = 0;            for (int j = 0; j <= len; j++)                if (isdigit(str[j]))                    cur = cur * 10 + str[j] - '0';                else G[tmp][cur] = 0,cur = 0;        }        dp[1][1] = 1;        G[1][1] = 0;        for (int i = 1; i <= n; i++)            for (int j = 1; j <= m; j++)                if (G[i][j])                     dp[i][j] = dp[i-1][j] + dp[i][j-1];                printf("%d\n",dp[n][m]);        if (t)            printf("\n");    }    return 0;}