题目链接:http://lightoj.com/volume_showproblem.php?problem=1281
题目给出n个城市m条已存在的道路,以及通过该道路需要的时间,然后就是k条可以修建的,但是由于预算的问题,最多只能修建d条道路,问这种情况下从1到n-1的最小时间花费。
不知道是被题目还是被自己逗了,,,真心没有读出来是单向边,wa得我都怀疑人生了。。。。
/***************************************** Author :Crazy_AC(JamesQi) Time :2015 File Name : *****************************************/ // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <iomanip> #include <sstream> #include <string> #include <stack> #include <queue> #include <deque> #include <vector> #include <map> #include <set> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <climits> using namespace std; #define MEM(x,y) memset(x, y,sizeof x) #define pk push_back typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> ii; typedef pair<ii,int> iii; const double eps = 1e-10; const int inf = 1 << 30; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; /********************************************************/ /**********************Point*****************************/ /********************************************************/ /* struct Point{ double x,y; Point(double x=0,double y=0):x(x),y(y){} }; typedef Point Vector; Vector operator + (Vector A,Vector B){ return Vector(A.x + B.x,A.y + B.y); } Vector operator - (Vector A,Vector B){//向量减法 return Vector(A.x - B.x,A.y - B.y); } Vector operator * (Vector A,double p){//向量数乘 return Vector(A.x * p,A.y * p); } Vector operator / (Vector A,double p){//向量除实数 return Vector(A.x / p,A.y / p); } int dcmp(double x){//精度正负、0的判断 if (fabs(x) < eps) return 0; return x < 0?-1:1; } bool operator < (const Point& A,const Point& B){//小于符号的重载 return A.x < B.x || (A.x == B.x && A.y < B.y); } bool operator == (const Point& A,const Point& B){//点重的判断 return dcmp(A.x - B.x) == 0&& dcmp(A.y - B.y) == 0; } double Dot(Vector A,Vector B){//向量的点乘 return A.x * B.x + A.y * B.y; } double Length(Vector A){//向量的模 return sqrt(Dot(A,A)); } double Angle(Vector A,Vector B){//向量的夹角 return acos(Dot(A,B) / Length(A) / Length(B)); } double Cross(Vector A,Vector B){//向量的叉积 return A.x * B.y - A.y * B.x; } double Area2(Point A,Point B,Point C){//三角形面积 return Cross(B - A,C - A); } Vector Rotate(Vector A,double rad){//向量的旋转 return Vector(A.x * cos(rad) - A.y * sin(rad),A.x * sin(rad) + A.y * cos(rad)); } Vector Normal(Vector A){//法向量 int L = Length(A); return Vector(-A.y / L,A.x / L); } double DistanceToLine(Point p,Point A,Point B){//p到直线AB的距离 Vector v1 = B - A,v2 = p - A; return fabs(Cross(v1,v2)) / Length(v1); } double DistanceToSegment(Point p,Point A,Point B){//p到线段AB的距离 if (A == B) return Length(p - A); Vector v1 = B - A, v2 = p - A,v3 = p - B; if (dcmp(Dot(v1,v2 < 0))) return Length(v2); else if (dcmp(Dot(v1,v3)) > 0) return Length(v3); else return DistanceToLine(p,A,B); } Point GetLineProjection(Point p,Point A,Point B){//投影 Vector v = B - A; return A + v * (Dot(v, p - A) / Dot(v, v)); } bool SegmentProperIntersection(Point A1,Point A2,Point B1,Point B2){//线段相交 double c1 = Cross(A2 - A1,B1 - A1),c2 = Cross(A2 - A1,B2 - A1); double c3 = Cross(B2 - B1,A1 - B1),c4 = Cross(B2 - B1,A2 - B1); return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0; } bool OnSegment(Point p,Point A,Point B){//点在线段上 return dcmp(Cross(A - p,B - p)) == 0 && dcmp(Dot(A - p,B - p)) < 0; } double PolygonArea(Point* p,int n){//多边形面积 double area = 0; for (int i = 1;i < n - 1;++i){ area += Cross(p[i] - p[0],p[i + 1] - p[0]); } return area / 2; } double PolygonArea(vector<Point> p){//多边形面积 Point O(0,0); double ret = 0; p.push_back(p[0]); for (int i = 0;i < p.size() - 1;++i){ ret += Cross(p[i] - O,p[i + 1] - O); } return ret / 2.0; } */ /********************************************************/ /**********************Line******************************/ /********************************************************/ /* struct Line{ Point p;//直线上任意一点 Vector v;//方向向量,它的左边就是半平面 double ang;//极角 Line(){} // Line(Point p,Vector v):p(p),v(v){} Line(Point A,Point B):p(A){ v = B - A; ang = atan2(v.y,v.x); } bool operator < (const Line& rhs)const{ return ang < rhs.ang; } }; */ const int maxn = 1e4 + 10; struct Edge{ int to, cost; bool newly; Edge(int to,int cost,int newly):to(to),cost(cost),newly(newly){} }; vector<Edge> G[maxn]; int n, m, k, d; int dis[maxn][12]; int solve(){ int st = 0, ed = n - 1; priority_queue<iii, vector<iii>, greater<iii> > que; que.push(iii(ii(0, 0), 0)); memset(dis, INF,sizeof dis); dis[0][0] = 0; while(!que.empty()){ iii tmp = que.top(); que.pop(); int u = tmp.second; int w = tmp.first.first; int cnt = tmp.first.second; if (u == ed) return dis[u][cnt]; if (w > dis[u][cnt]) continue; if (cnt > d) continue; int size = G[u].size(); for (int i = 0;i < size;++i){ int v = G[u][i].to; int cost = G[u][i].cost; bool is = G[u][i].newly; if (is){ if (cnt + 1 <= d && dis[v][cnt + 1] > dis[u][cnt] + cost){ dis[v][cnt + 1] = dis[u][cnt] + cost; que.push(iii(ii(dis[v][cnt + 1],cnt + 1), v)); } }else{ if (dis[v][cnt] > dis[u][cnt] + cost){ dis[v][cnt] = dis[u][cnt] + cost; que.push(iii(ii(dis[v][cnt], cnt), v)); } } } } int ans = INF; // for (int i = 0;i <= d;++i) // ans = min(ans, dis[n-1][i]); return ans; } int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int t, icase = 0; scanf("%d",&t); while(t--){ scanf("%d%d%d%d",&n,&m,&k,&d); for (int i = 0;i < n;++i) G[i].clear(); int u, v, w; for (int i = 0;i < m;++i){ scanf("%d%d%d",&u,&v,&w); G[u].push_back(Edge(v, w, false)); // G[v].push_back(Edge(u, w, false)); } for (int i = 0;i < k;++i){ scanf("%d%d%d",&u,&v,&w); G[u].push_back(Edge(v, w, true)); // G[v].push_back(Edge(u, w, true)); } int ans = solve(); printf("Case %d: ", ++icase); if (ans == INF) puts("Impossible"); else printf("%d\n", ans); } return 0; }