A Walk Through the Forest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5421 Accepted Submission(s):
1988
Problem Description
Jimmy experiences a lot of stress at work these days,
especially since his accident made working difficult. To relax after a hard day,
he likes to walk home. To make things even nicer, his office is on one side of a
forest, and his house is on the other. A nice walk through the forest, seeing
the birds and chipmunks is quite enjoyable.
The forest is beautiful, and
Jimmy wants to take a different route everyday. He also wants to get home before
dark, so he always takes a path to make progress towards his house. He considers
taking a path from A to B to be progress if there exists a route from B to his
home that is shorter than any possible route from A. Calculate how many
different routes through the forest Jimmy might take.
especially since his accident made working difficult. To relax after a hard day,
he likes to walk home. To make things even nicer, his office is on one side of a
forest, and his house is on the other. A nice walk through the forest, seeing
the birds and chipmunks is quite enjoyable.
The forest is beautiful, and
Jimmy wants to take a different route everyday. He also wants to get home before
dark, so he always takes a path to make progress towards his house. He considers
taking a path from A to B to be progress if there exists a route from B to his
home that is shorter than any possible route from A. Calculate how many
different routes through the forest Jimmy might take.
Input
Input contains several test cases followed by a line
containing 0. Jimmy has numbered each intersection or joining of paths starting
with 1. His office is numbered 1, and his house is numbered 2. The first line of
each test case gives the number of intersections N, 1 < N ≤ 1000, and the
number of paths M. The following M lines each contain a pair of intersections a
b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between
intersection a and a different intersection b. Jimmy may walk a path any
direction he chooses. There is at most one path between any pair of
intersections.
containing 0. Jimmy has numbered each intersection or joining of paths starting
with 1. His office is numbered 1, and his house is numbered 2. The first line of
each test case gives the number of intersections N, 1 < N ≤ 1000, and the
number of paths M. The following M lines each contain a pair of intersections a
b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between
intersection a and a different intersection b. Jimmy may walk a path any
direction he chooses. There is at most one path between any pair of
intersections.
Output
For each test case, output a single integer indicating
the number of different routes through the forest. You may assume that this
number does not exceed 2147483647
the number of different routes through the forest. You may assume that this
number does not exceed 2147483647
Sample Input
5 6
1 3 2
1 4 2
3 4 3
1 5 12
4 2 34
5 2 24
7 8
1 3 1
1 4 1
3 7 1
7 4 1
7 5 1
6 7 1
5 2 1
6 2 1
Sample Output
2
4
思路：
用dis求最短路径，用dfs求最短路径的数量
代码：
#include<stdio.h>
#include<stdlib.h>
#include<string.h> #define INF 0xfffffff int map[][], disk[], pathnum[];
int n, m; int getmin(int x, int y){
return x > y ? y : x;
} int dis(){
int i, j, visit[], idmin, min;
memset(visit, , sizeof(visit));
for(i = ; i <= n; i ++){
disk[i] = map[][i];
}
disk[] = ;
for(i = ; i <= n; i ++){
idmin = ;
min = INF;
for(j = ; j <= n; j ++){
if(!visit[j] && disk[j] < min){
min = disk[j];
idmin = j;
}
}
visit[idmin] = ;
for(j = ; j <= n; j ++){
if(!visit[j]){
disk[j] = getmin(disk[j], map[idmin][j] + disk[idmin]);
}
}
}
return ;
} int dfs(int start){
int sum, i;
if(start == ){
return ;
}
if(pathnum[start] != -){
return pathnum[start];
}
sum = ;
for(i = ; i <= n; i ++){
if(map[i][start] != INF && map[i][start] == disk[start] - disk[i]){
sum += dfs(i);
}
}
pathnum[start] = sum;
return pathnum[start];
} int main(){
int x, y, d, i, j;
while(scanf("%d", &n) && n){
scanf("%d", &m);
for(i = ; i < ; i ++){
for(j = ; j < ; j ++){
map[i][j] = INF;
}
}
//printf("%d\n", map[1][1]);
for(i = ; i < m; i ++){
scanf("%d %d %d", &x, &y, &d);
map[x][y] = map[y][x] = d;
}
dis();
//printf("%d\n", disk[1]);
memset(pathnum, -, sizeof(pathnum));
//printf("%d\n", pathnum[1]);
printf("%d\n", dfs());
}
return ;
}