（最短路深入）POJ 3463 - Sightseeing

Posted 2020-09-08 tak_fate

题意：

给一个有向图，计算最短路和比最短路少1的路的条数的和。

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分析：

这题真的写死我了。

因为之前很少接触最短路问题，所谓偶尔遇到一次也是套的模板，根本没有细细思考过dijsktra算法。所以栽在了这题上。

这题就是求最短路和次短路。

核心思想在于修改最短路松弛的条件，并且每个节点同时维护最短路和次短路。

很多博主写的很详细，我也不多说了，只是写个博文记录一下自己有多渣，在学习算法的道路上自己真的思考的不够多，也不够努力。

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代码：

  1 #include <set>
  2 #include <map>
  3 #include <list>
  4 #include <cmath>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <string>
 10 #include <cctype>
 11 #include <cstdio>
 12 #include <cstring>
 13 #include <cstdlib>
 14 #include <iostream>
 15 #include <algorithm>
 16 // #include <unordered_map>
 17 
 18 using namespace std;
 19 
 20 typedef long long ll;
 21 typedef unsigned long long ull;
 22 typedef pair<int, int> pii;
 23 typedef pair<ull, ull> puu;
 24 
 25 #define inf (0x3f3f3f3f)
 26 #define lnf (0x3f3f3f3f3f3f3f3f)
 27 #define eps (1e-9)
 28 #define fi first
 29 #define se second
 30 
 31 bool sgn(double a, string select, double b) {
 32     if(select == "==")return fabs(a - b) < eps;
 33     if(select == "!=")return fabs(a - b) > eps;
 34     if(select == "<")return a - b < -eps;
 35     if(select == "<=")return a - b < eps;
 36     if(select == ">")return a - b > eps;
 37     if(select == ">=")return a - b > -eps;
 38 }
 39 
 40 
 41 //--------------------------
 42 
 43 const ll mod = 1000000007;
 44 const int maxn = 100010;
 45 
 46 //use to bfs
 47 struct Node {
 48     int v, c;
 49     int kind;
 50     Node(int _v = 0, int _c = 0, int _kind = 0) {
 51         v = _v, c = _c, kind = _kind;
 52     }
 53     bool operator<(const Node &a)const {
 54         return c > a.c;
 55     }
 56 };
 57 
 58 //first is ‘v‘ , second is ‘cost‘
 59 vector<pii> edge[maxn];
 60 bool vis[maxn][2];
 61 int dist[maxn][2];
 62 int sav[maxn][2];
 63 
 64 void addedge(int u, int v, int w) {
 65     edge[u].push_back(make_pair(v, w));
 66 }
 67 
 68 
 69 
 70 // id from 1
 71 int  dijkstra_heap(int n, int start, int stop) {
 72     memset(vis, 0, sizeof(vis));
 73     memset(sav, 0, sizeof(sav));
 74     for(int i = 1; i <= n; i++) dist[i][0] = dist[i][1] = inf;
 75     priority_queue<Node> que;
 76     while(!que.empty())que.pop();
 77     dist[start][0] = 0;
 78     sav[start][0] = 1;
 79     que.push(Node(start, 0, 0));
 80     Node tmp;
 81     while(!que.empty()) {
 82         tmp = que.top();
 83         que.pop();
 84         int u = tmp.v;
 85         int kind = tmp.kind;
 86         if(vis[u][kind])continue;
 87         vis[u][kind] = true;
 88         for(int i = 0; i < edge[u].size(); i++) {
 89             int v = edge[u][i].fi;
 90             int cost = edge[u][i].se;
 91             if(dist[u][kind] + cost < dist[v][0]) {
 92                 dist[v][1] = dist[v][0];
 93                 sav[v][1] = sav[v][0];
 94                 dist[v][0] = dist[u][kind] + cost;
 95                 sav[v][0] = sav[u][kind];
 96                 que.push(Node(v, dist[v][0], 0));
 97                 que.push(Node(v, dist[v][1], 1));
 98             } else if(dist[u][kind] + cost == dist[v][0]) {
 99                 sav[v][0] += sav[u][kind];
100             } else if(dist[u][kind] + cost < dist[v][1]) {
101                 dist[v][1] = dist[u][kind] + cost;
102                 sav[v][1] = sav[u][kind];
103                 que.push(Node(v, dist[v][1], 1));
104             } else if(dist[u][kind] + cost == dist[v][1]) {
105                 sav[v][1] += sav[u][kind];
106             }
107         }
108     }
109     if(dist[stop][0] + 1 == dist[stop][1])return sav[stop][0] + sav[stop][1];
110     return sav[stop][0];
111 }
112 
113 int s, t;
114 int n, m;
115 
116 
117 void solve() {
118     int kase;
119     scanf("%d", &kase);
120     while(kase--) {
121         for(int i = 1; i <= n; i++) {
122             edge[i].clear();
123         }
124         scanf("%d%d", &n, &m);
125         int u, v, w;
126         for(int i = 0; i < m; i++) {
127             scanf("%d%d%d", &u, &v, &w);
128             addedge(u, v, w);
129         }
130         scanf("%d%d", &s, &t);
131         int ans = dijkstra_heap(n, s, t);
132         printf("%d\n", ans );
133     }
134 
135 }
136 
137 int main() {
138 
139 #ifndef ONLINE_JUDGE
140     freopen("1.in", "r", stdin);
141     freopen("1.out", "w", stdout);
142 #endif
143     // iostream::sync_with_stdio(false);
144     solve();
145     return 0;
146 }