A Walk Through the Forest

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

2
4

题意：若a到终点的距离大于b到终点的距离，而且a跟b联通，那么可以从a经过b到达终点，求从起点到终点的路有多少条。
题解：SPFA求终点到各个点的最短路径，DFS搜索可行路的条数。

#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 1002
#define inf 0x3f3f3f3f
using std::queue;

int map[maxn][maxn];
int dist[maxn], dp[maxn];
bool vis[maxn];

void SPFA(int s, int n)
{
	int i, u, tmp;
	for(i = 0; i <= n; ++i){
		dist[i] = inf; vis[i] = false;
	}
	queue<int> Q;
	u = s; vis[u] = 1; dist[u] = 0;
	Q.push(u);
	while(!Q.empty()){
		u = Q.front(); Q.pop();
		vis[u] = false;
		for(i = 1; i <= n; ++i){
			if(map[u][i] == inf) continue;
			tmp = map[u][i] + dist[u];
			if(tmp < dist[i]){
				dist[i] = tmp;
				vis[i] = true;
				Q.push(i);
			}
		}
	}
}

int DFS(int s, int n)
{
	if(dp[s]) return dp[s];
	if(s == 2) return 1;
	int i, ans = 0;
	for(i = 1; i <= n; ++i)
		if(map[s][i] != inf && dist[i] < dist[s]){
			ans += DFS(i, n);
		}
	return dp[s] = ans;
}

int main()
{
	int n, m, u, v, d, i, j;
	while(scanf("%d%d", &n, &m) == 2){
		for(i = 0; i <= n; ++i){
			for(j = 0; j <= n; ++j)
				map[i][j] = inf;
			map[i][i] = 0;
		}
		for(i = 0; i < m; ++i){
			scanf("%d%d%d", &u, &v, &d);
			map[u][v] = map[v][u] = d;
		}
		SPFA(2, n);
		for(i = 0; i <= n; ++i) dp[i] = 0;
		printf("%d\n", DFS(1, n));
	}
	return 0;
}

Dijkstra 实现：

#include <stdio.h>
#include <string.h>
#define maxn 1002
#define inf 0x3f3f3f3f

int map[maxn][maxn];
int dist[maxn], dp[maxn];
bool vis[maxn];

int getNext(int n)
{
	int u = -1, tmp = inf, i;
	for(i = 1; i <= n; ++i)
		if(!vis[i] && tmp > dist[i]){
			tmp = dist[i]; u = i;
		}
	return u;
}

void Dijkstra(int s, int n)
{
	int i, u, tmp;
	for(i = 0; i <= n; ++i){
		dist[i] = inf; vis[i] = false;
	}
	u = s; dist[u] = 0;
	while(u != -1){
		for(i = 1; i <= n; ++i){
			if(map[u][i] == inf) continue;
			tmp = map[u][i] + dist[u];
			if(tmp < dist[i]) dist[i] = tmp;
		}
		vis[u] = true;
		u = getNext(n);
	}
}

int DFS(int s, int n)
{
	if(dp[s]) return dp[s];
	if(s == 2) return 1;
	int i, ans = 0;
	for(i = 1; i <= n; ++i)
		if(map[s][i] != inf && dist[i] < dist[s]){
			ans += DFS(i, n);
		}
	return dp[s] = ans;
}

int main()
{
	int n, m, u, v, d, i, j;
	while(scanf("%d%d", &n, &m) == 2){
		for(i = 0; i <= n; ++i){
			for(j = 0; j <= n; ++j)
				map[i][j] = inf;
			map[i][i] = 0;
		}
		for(i = 0; i < m; ++i){
			scanf("%d%d%d", &u, &v, &d);
			map[u][v] = map[v][u] = d;
		}
		Dijkstra(2, n);
		for(i = 0; i <= n; ++i) dp[i] = 0;
		printf("%d\n", DFS(1, n));
	}
	return 0;
}

HDU1142 A Walk Through the Forest 【SPFA】+【记忆化搜索】,布布扣,bubuko.com

HDU1142 A Walk Through the Forest 【SPFA】+【记忆化搜索】

原文：http://blog.csdn.net/chang_mu/article/details/38554347