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;
}