UVa 10917 - Walk Through the Forest（Dijkstra + DFS)

题意

小明要从办公室回家，当从A到家里存在一条路比B到家的任何一条路都短的话，就走A。问他有几种走法。

思路

意思就是A到终点的最短路比B的最短路短的时候可以走A。

先对终点进行一次Dijkstra，求出每个点到终点的距离，然后DFS。

代码

#include<bits/stdc++.h>

#define LL long long

#define lowbit(x) ((x) & (-x))

#define MP(a, b) make_pair(a, b)

const int MAXN = 1000 + 5;

const int INF = 0x3f3f3f3f;

using namespace std;

typedef pair<int, int> pii;

typedef vector<int> vei;

typedef vector<pair<int, int> >veii;

typedef vector<int>::iterator viti;

typedef vector<pii>::iterator vitii;

typedef priority_queue<pii, vector<pii>, greater<pii> >pquii;

pquii pqu;

int dis[MAXN], dp[MAXN];

veii mp[MAXN];

void Dijkstra()

{

memset(dis, INF, sizeof dis);

dis[2] = 0;

pqu.push(MP(dis[2], 2));

while (!pqu.empty())

{

pii u = pqu.top(); pqu.pop();

int x = u.second;

if (dis[x] != u.first) continue;

for (vitii it = mp[x].begin(); it != mp[x].end(); it++)

{

int a = it->first, b = it->second;

if (dis[a] > dis[x] + b)

{

dis[a] = dis[x] + b;

pqu.push(MP(dis[a], a));

}

}

}

}

int DFS(int x)

{

if (dp[x] != -1) return dp[x];

if (x == 2) return 1;

int sum = 0;

for (vitii it = mp[x].begin(); it != mp[x].end(); it++)

if (dis[it->first] < dis[x]) sum += DFS(it->first);

return dp[x] = sum;

}

int main()

{

//freopen("input.txt", "r", stdin);

int nroad, njit, i, j;

while (scanf("%d", &njit), njit)

{

scanf("%d", &nroad);

for (i = 0; i <= njit; i++) mp[i].clear();

for (i = 0; i < nroad; i++)

{

int a, b, c;

scanf("%d%d%d", &a, &b, &c);

mp[a].push_back(MP(b, c)); mp[b].push_back(MP(a, c));

}

Dijkstra();

memset(dp, -1, sizeof dp);

printf("%d\n", DFS(1));

}

return 0;

}