（最短路深入）POJ 3463
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。
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。
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 ", 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 }