POJ 3463 Sightseeing (最短路&次短路條數問題)
阿新 • • 發佈:2019-01-04
題意:給一個有向圖,求從s到f 的最短路+最短路-1的條數。 有重邊。
分析: 程式碼是根據挑戰的次短路改編的。 具體請看程式碼。
#include<iostream> #include<algorithm> #include<cstdio> #include<cmath> #include<cstring> #include<string> #include<vector> #include<cctype> #include<set> #include<map> #include<queue> #include<stack> #include<iomanip> #include<sstream> #include<limits> #define ll long long #define inf 0x3f3f3f3f using namespace std; const ll INF = 1e18; const int maxn = 2e5+10; const ll MOD = 1000000007; const double EPS = 1e-10; const double Pi = acos(-1.0); struct edge{int to,cost;}; //typedef pair<int,int>P; //1 dist 2 u struct P{ int first,second,z; //z 標誌是最短路還是次短路 P(int a=0,int b=0,int c=0):first(a),second(b),z(c){} bool operator < (const P &rhs)const{return first>rhs.first;} }; int V; vector<edge>G[maxn]; int d[maxn],d2[maxn],cnt1[maxn],cnt2[maxn],vis[maxn][2]; //d cnt1 最短路長度及數量 d2 cnt2 次短路長度及數量 vis代表有沒訪問過 void dij(int s) { priority_queue<P >que; fill(d,d+V+1,inf); fill(d2,d2+V+1,inf); d[s] = 0; cnt1[s] = 1; que.push(P(0,s,0)); while(!que.empty()) { P p = que.top(); que.pop(); int v = p.second, z = p.z; if (d2[v] < p.first || vis[v][z]) continue; vis[v][z] = 1; //一定要記錄是否訪問過,不然會出錯,因為是單調佇列。 for(int i = 0; i < G[v].size();i++) { edge e = G[v][i]; int temp = p.first + e.cost; if (d[e.to] > temp) // 比最短路短 { d2[e.to] = d[e.to]; d[e.to] = temp; cnt2[e.to] = cnt1[e.to]; cnt1[e.to] = z==0? cnt1[v]:cnt2[v]; //要弄清楚等於哪個。 que.push(P(d[e.to] ,e.to,0)); que.push(P(d2[e.to] ,e.to,1)); }else if (d[e.to] == temp) cnt1[e.to] += z==0? cnt1[v]:cnt2[v]; // 等於最短路 else if (temp < d2[e.to]) // 比次短路短 { d2[e.to] = temp; cnt2[e.to] = z==0? cnt1[v]:cnt2[v]; que.push(P(d2[e.to] ,e.to,1)); }else if (temp == d2[e.to]) cnt2[e.to] += z==0? cnt1[v]:cnt2[v]; // 等於次短路 } } } int main(){ #ifdef LOCAL freopen("C:\\Users\\lanjiaming\\Desktop\\acm\\in.txt","r",stdin); //freopen("output.txt","w",stdout); #endif //ios_base::sync_with_stdio(0); int T,n,m; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(int i = 0; i <= n; i++) G[i].clear(); V = n; for(int i = 0; i < m; i++) { int a,b,l; scanf("%d%d%d",&a,&b,&l); G[a].push_back(edge{b,l}); } int s,f; scanf("%d%d",&s,&f); memset(cnt1,0,sizeof cnt1); memset(cnt2,0,sizeof cnt2); memset(vis,0,sizeof vis); dij(s); if (d[f] == d2[f]-1) cnt1[f] += cnt2[f]; printf("%d\n",cnt1[f]); } return 0; }