UVA - 10917 Walk Through the Forest(최단로+DP)

Description

Problem C: A Walk Through the Forest

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.

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.

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.

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
0

Output for Sample Input

2
4

제목: Jimmy는 매일 다른 길을 따라 가려고 한다. 그리고 그는 다음과 같은 조건을 충족시키는 길을 따라갈 수 밖에 없다(A, B): B에서 집으로 돌아가는 경로가 존재한다. 그래서 A에서 집으로 돌아가는 경로보다 짧다. 너의 임무는 몇 개의 다른 경로가 있는지 계산하는 것이다.
사고방식: 그래서 집으로 돌아가는 가장 짧은 경로를 구한다. 그 다음에dist[A]>dist[B]의 모든 가능한 경로를 만족시키기 위해dist[A]>dist[B]를 통해 우리는 새로운 그림을 만들 수 있다. 제목은 시작점에서 끝까지의 경로 개수를 구하는 것이다.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int MAXN = 1010;
const int INF = 0x3f3f3f3f;

int dp[MAXN], dist[MAXN], cost[MAXN][MAXN];
int vis[MAXN];
int n, m, s, e, w;

void Dijkstra() {
	int Min, cnt;
	for (int i = 1; i <= n; i++)
		dist[i] = cost[2][i];
	vis[2] = 1;
	for (int i = 1; i <= n; i++) {
		Min = INF;
		for (int j = 1; j <= n; j++)
			if (!vis[j] && dist[j] < Min) {
				Min = dist[j];
				cnt = j;
			}
		vis[cnt] = 1;
		if (Min == INF)
			break;
		for (int k = 1; k <= n; k++)
			if (!vis[k] && dist[k] > dist[cnt]+cost[k][cnt])
				dist[k] = dist[cnt] + cost[k][cnt];
	}
}

int dfs(int cur) {
	if (dp[cur] > 0)
		return dp[cur];
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		if (dist[cur] > dist[i] && cost[cur][i] != INF)
			ans += dfs(i);
	}
	return dp[cur] = ans;
}

int main() {
	while (scanf("%d", &n) != EOF && n) {
		scanf("%d", &m);
		memset(cost, INF, sizeof(cost));
		memset(dist, INF, sizeof(dist));
		memset(vis, 0, sizeof(vis));
		memset(dp, 0, sizeof(dp));
		for (int i = 0; i <= n; i++)
			cost[i][i] = 0;
		for (int i = 0; i < m; i++) {
			scanf("%d%d%d", &s, &e, &w);
			cost[s][e] = cost[e][s] = w;
		}
		Dijkstra();
		dp[2] = 1;
		int ans = dfs(1);
		printf("%d
", ans);
	}	
	return 0;
}