输入处理懒得写了，直接复制了，23333
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。
#include <stdio.h>
#include <string.h>
const int N = 1005;
int n, m, dp[N][N], fb[N][N];
int dfs(int x,int y)
{
if(dp[x][y]!=-1)
return dp[x][y];
if(x==n&&y==m)
return dp[x][y]=1;
dp[x][y]=0;
if(x+1<=n&&!fb[x+1][y])
dp[x][y]+=dfs(x+1,y);
if(y+1<=m&&!fb[x][y+1])
dp[x][y]+=dfs(x,y+1);
return dp[x][y];
}
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 {
fb[k][num] = 1;
num = 0;
}
}
}
void read() {
int r;
char str[N];
memset(fb, 0, sizeof(fb));
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%d", &r);
gets(str);
handle(r, str);
}
}
int main() {
int cas;
scanf("%d", &cas);
while (cas--) {
read();
memset(dp,-1,sizeof(dp));
printf("%d\n", dfs(1,1));
if (cas) printf("\n");
}
return 0;
}
| Time Limit:3000MS | Memory Limit:Unknown | 64bit IO Format:%lld & %llu |
Description
Walking on the Safe Side
| Walking on the Safe Side |
Square City is a very easy place for people to walk around. The two-way streets run North-South or East-West dividing the city into regular blocks. Most street intersections are safe for pedestrians to cross. In some of them, however, crossing is not safe and
pedestrians are forced to use the available underground passages. Such intersections are avoided by walkers. The entry to the city park is on the North-West corner of town, whereas the railway station is on the South-East corner.
Suppose you want to go from the park to the railway station, and do not want to walk more than the required number of blocks. You also want to make your way avoiding the underground passages, that would introduce extra delay. Your task is to determine the number
of different paths that you can follow from the park to the station, satisfying both requirements.
The example in the picture illustrates a city with 4 E-W streets and 5 N-S streets. Three intersections are marked as unsafe. The path from the park to the station is 3 + 4 = 7 blocks long and there are 4 such paths that avoid the underground passages.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The first line of the input contains the number of East-West streets
W and the number of North-South streets N. Each one of the following
W lines starts with the number of an East-West street, followed by zero or more numbers of the North-South crossings which are unsafe. Streets are numbered from 1.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
The number of different minimal paths from the park to the station avoiding underground passages.
Sample Input
1 4 5 1 2 2 3 3 5 4
Sample Output
4
Cristina Ribeiro, MIUP'2001
Source
Volume 5. Dynamic Programming
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 4. Graph :: Special Graph - DAG ::
Counting Paths in DAG
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Graph :: Special Graph (Directed Acyclic Graph) ::
Counting Paths in DAG
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Graph :: Special Graph (Directed Acyclic Graph) ::
Counting Paths in DAG