UVA10917 Walk Through the Forest

题目大意：Jimmy下班后决定每天沿着一条不同的路径回家，欣赏不同的风景。他打算只沿着满足如下条件的（A,B）道路走：存在一条从B出发回家的路，比所有从A出发回家的路径都短。你的任务是计算一共有多少条不同的回家路径。其中公司的编号为1，家的编号为2.

题解：算出每个点到2的最短路，对于每条边(i,j)，“存在一条从B出发回家的路，比所有从A出发回家的路径都短”，即d[B] < d[A]，说明能从A走到B，于是建新图A->B。不难发现是一个DAG（因为有环就会出现逻辑矛盾）。建反图dp即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <cmath>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
template<class T>
inline void swap(T &a, T &b)
{
T tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
x = 0;char ch = getchar(), c = ch;
while(ch < '0' || ch > '9') c = ch, ch = getchar();
while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;
const int MAXN = 1000 + 10;
struct Edge
{
int u,v,w,nxt;
Edge(int _u, int _v, int _w, int _nxt){u = _u;v = _v;w = _w;nxt = _nxt;}
Edge(){}
}edge1[MAXN * MAXN], edge2[MAXN * MAXN];
int head1[MAXN], cnt1, head2[MAXN], cnt2;
inline void insert1(int a, int b, int c){edge1[++cnt1] = Edge(a,b,c,head1[a]), head1[a] = cnt1;}
inline void insert2(int a, int b, int c){edge2[++cnt2] = Edge(a,b,c,head2[a]), head2[a] = cnt2;}
struct Node
{
int u,w;
Node(int _u, int _w){u = _u;w = _w;}
Node(){}
};
struct cmp
{
bool operator()(Node a, Node b)
{
return a.w > b.w;
}
};
std::priority_queue<Node, std::vector<Node>, cmp> q;
int n,m,d[MAXN],vis[MAXN],tmp1,tmp2,tmp3,dp[MAXN];
void dij(int S)
{
memset(d, 0x3f, sizeof(d)), d[S] = 0, memset(vis, 0, sizeof(vis)), q.push(Node(S, 0));
while(q.size())
{
Node now = q.top();q.pop();
if(vis[now.u]) continue;vis[now.u] = 1;
for(int pos = head1[now.u];pos;pos = edge1[pos].nxt)
{
int v = edge1[pos].v;
if(vis[v]) continue;
if(d[v] > d[now.u] + edge1[pos].w) d[v] = d[now.u] + edge1[pos].w, q.push(Node(v, d[v]));
}
}
}
int Dp(int x)
{
if(vis[x]) return dp[x];
for(int pos = head2[x];pos;pos = edge2[pos].nxt)
{
int v = edge2[pos].v;
dp[x] += Dp(v);
}
vis[x] = 1;
return dp[x];
}
int main()
{
while(scanf("%d", &n) != EOF && n)
{
read(m), memset(head1, 0, sizeof(head1)), memset(head2, 0, sizeof(head2)), cnt1 = 0, cnt2 = 0;
for(int i = 1;i <= m;++ i) read(tmp1), read(tmp2), read(tmp3), insert1(tmp1, tmp2, tmp3), insert1(tmp2, tmp1, tmp3);
dij(2);
for(int i = 1;i <= cnt1;++ i)
if(d[edge1[i].v] < d[edge1[i].u])
insert2(edge1[i].v, edge1[i].u, edge1[i].w);
memset(vis, 0, sizeof(vis)), memset(dp, 0, sizeof(dp)), vis[1] = 0, dp[1] = 1;
printf("%d\n", Dp(2));
}
return 0;
}

UVA10917