There are nn planets in the MOT galaxy, and each planet has a unique number from 1 \sim n1∼n. Each planet is connected to other planets through some transmission channels. There are mm transmission channels in the galaxy. Each transmission channel connects two different planets, and each transmission channel has a length.
The residents of the galaxy complete the interplanetary voyage by spaceship. Each spaceship has a level. The spacecraft can be upgraded several times. It can only be upgraded 11 level each time, and the cost is cc. Each upgrade will increase the transmission distance by dd and the number of transmissions channels by ee to the spacecraft. The spacecraft can only pass through channels that are shorter than or equal to its transmission distance. If the number of transmissions is exhausted, the spacecraft can no longer be used.
Alice initially has a 00-level spacecraft with transmission distance of 00 and transmission number of 00. Alice wants to know how much it costs at least, in order to transfer from planet 11 to planet nn.
Input
Each test file contains a single test case. In each test file:
The first line contains nn, mm, indicating the number of plants and the number of transmission channels
The second line contains cc, dd, ee, representing the cost, the increased transmission distance, and the increased number of transmissions channels of each upgrade, respectively.
Next mm lines, each line contains u,v,wu,v,w, meaning that there is a transmission channel between uu and vv with a length of ww.
(2 \le n\le 10^5, n - 1 \le m \le 10^5,1 \le u,v \le n ,1 \le c,d,e,w \le 10^5)(2≤n≤105,n−1≤m≤105,1≤u,v≤n,1≤c,d,e,w≤105)
(The graph has no self-loop , no repeated edges , and is connected)
Output
Output a line for the minimum cost. Output -1−1 if she can't reach.
样例输入
5 7 1 1 1 1 2 1 1 3 5 1 4 1 2 3 2 2 4 5 3 4 3 3 5 5
样例输出
5
二分+最短路。
代码:
1 //M-二分+最短路dijkstra-二分过程中保存结果，因为二分完最后的不一定是结果。 2 #include<bits/stdc++.h> 3 using namespace std; 4 typedef long long ll; 5 const int maxn=1e5+10; 6 const int inf=0x3f3f3f3f; 7 #define pii pair<int,int> 8 #define mp make_pair 9 #define fi first 10 #define se second 11 12 namespace IO{ 13 char buf[1<<15],*S,*T; 14 inline char gc(){ 15 if (S==T){ 16 T=(S=buf)+fread(buf,1,1<<15,stdin); 17 if (S==T)return EOF; 18 } 19 return *S++; 20 } 21 inline int read(){ 22 int x; bool f; char c; 23 for(f=0;(c=gc())<'0'||c>'9';f=c=='-'); 24 for(x=c^'0';(c=gc())>='0'&&c<='9';x=(x<<3)+(x<<1)+(c^'0')); 25 return f?-x:x; 26 } 27 inline long long readll(){ 28 long long x;bool f;char c; 29 for(f=0;(c=gc())<'0'||c>'9';f=c=='-'); 30 for(x=c^'0';(c=gc())>='0'&&c<='9';x=(x<<3)+(x<<1)+(c^'0')); 31 return f?-x:x; 32 } 33 } 34 using IO::read; 35 using IO::readll; 36 37 38 int n,m; 39 int c,d,e; 40 int head[maxn<<1],cnt; 41 int dis[maxn],pathmin,edgemin; 42 bool vis[maxn]; 43 priority_queue<pii,vector<pii>,greater<pii> > q; 44 45 struct Edge{ 46 int u,v,w; 47 }path[maxn]; 48 49 struct node{ 50 int to,next,w; 51 }edge[maxn<<1]; 52 53 void init() 54 { 55 memset(head,-1,sizeof head); 56 memset(edge,0,sizeof edge); 57 memset(vis,0,sizeof vis); 58 memset(dis,inf,sizeof dis); 59 cnt=0; 60 } 61 62 void add(int u,int v,int w) 63 { 64 edge[cnt].to=v; 65 edge[cnt].w=w; 66 edge[cnt].next=head[u]; 67 head[u]=cnt++; 68 } 69 70 void dijkstra(int s) 71 { 72 dis[s]=0; 73 q.push(mp(0,s)); 74 while(!q.empty()){ 75 int u=q.top().se;q.pop(); 76 vis[u]=true; 77 for(int i=head[u];~i;i=edge[i].next){ 78 int v=edge[i].to; 79 int w=edge[i].w; 80 if(dis[v]>dis[u]+w){ 81 dis[v]=dis[u]+w; 82 q.push(mp(dis[v],v)); 83 } 84 } 85 } 86 } 87 88 void solve() 89 { 90 pathmin=inf;edgemin=inf; 91 int l=1,r=1e5; 92 while(l<=r){ 93 int mid=(l+r)>>1; 94 init(); 95 for(int i=1;i<=m;i++){ 96 if(path[i].w<=mid){ 97 add(path[i].u,path[i].v,1); 98 add(path[i].v,path[i].u,1); 99 } 100 } 101 dijkstra(1); 102 if(dis[n]<=mid){ 103 r=mid-1; 104 if(edgemin>mid){ 105 edgemin=mid; 106 pathmin=dis[n]; 107 } 108 else if(edgemin==mid){ 109 pathmin=min(pathmin,dis[n]); 110 } 111 } 112 else{ 113 l=mid+1; 114 } 115 } 116 } 117 118 int main() 119 { 120 n=read();m=read(); 121 // scanf("%d%d",&n,&m); 122 c=read();d=read();e=read(); 123 // scanf("%d%d%d",&c,&d,&e); 124 for(int i=1;i<=m;i++){ 125 path[i].u=read();path[i].v=read();path[i].w=read(); 126 // scanf("%d%d%d",&path[i].u,&path[i].v,&path[i].w); 127 } 128 solve();//路径的最大边权 129 edgemin=edgemin/d+(edgemin%d==0? 0:1); 130 pathmin=pathmin/e+(pathmin%e==0? 0:1); //路径条数 131 printf("%lld\n",1ll*max(edgemin,pathmin)*c); 132 } 133 */ 134 /* 135 136 5 7 137 1 1 1 138 1 2 3 139 1 3 4 140 1 4 4 141 2 3 2 142 2 4 5 143 3 4 1 144 3 5 3 145 146 3 147 148 149 150 /* 151 //M-HY学长的代码orz 152 //二分+最短路spfa 153 #include<bits/stdc++.h> 154 #define INF 0x3f3f3f3f 155 using namespace std; 156 157 typedef long long ll; 158 const int N = 200010, mod = 1e9+7, M = 300005; 159 160 ll addCost, addDist, addNum; 161 162 int n, m, fa[N], head[N], tot2; 163 pair<ll, pair<int ,int > > e1[M]; 164 struct Edge{ int to, next; ll w; } e[M]; 165 void Init() { 166 tot2=0; 167 memset(head, -1, sizeof(head)); 168 } 169 void Addedge(int u, int v, ll w) { 170 e[tot2]={v, head[u], w}; head[u]=tot2++; 171 e[tot2]={u, head[v], w}; head[v]=tot2++; 172 } 173 174 queue<int > Q; 175 bool inQ[N]; 176 ll dis[N]; 177 bool spfa(int mid) 178 { 179 memset(inQ, false, sizeof(inQ)); 180 memset(dis, INF, sizeof(dis)); 181 dis[1] = 0; 182 Q.push(1); 183 inQ[1] = true; 184 ll mostDist = addDist * mid; 185 ll mostNum = addNum * mid; 186 while (!Q.empty()) 187 { 188 int u = Q.front(); //cout << u << " " << dis[u] << " ---- \n"; 189 Q.pop(); 190 for (int i = head[u]; i != -1; i = e[i].next) 191 { 192 if (e[i].w > mostDist) 193 continue; 194 if (dis[e[i].to] > dis[u] + 1) 195 { 196 dis[e[i].to] = dis[u] + 1; 197 if (!inQ[e[i].to]) 198 { 199 Q.push(e[i].to); 200 inQ[e[i].to] = true; 201 } 202 } 203 } 204 inQ[u] = false; 205 } 206 return dis[n] != INF && dis[n] <= mostNum; 207 } 208 209 int main() 210 { 211 scanf("%d %d", &n, &m); 212 scanf("%lld %lld %lld", &addCost, &addDist, &addNum); 213 Init(); 214 for (int i = 1; i <= m; ++ i) 215 { 216 scanf("%d %d %lld", &e1[i].second.first, &e1[i].second.second, &e1[i].first); 217 Addedge(e1[i].second.first, e1[i].second.second, e1[i].first); 218 } 219 220 //cout << 1111 << endl; 221 // 二分 222 int l = 0, r = N, mid, ans = INF; 223 while (l <= r) 224 { 225 mid = (l + r) / 2; 226 if (spfa(mid)) 227 ans = mid, r = mid -1; 228 else 229 l = mid + 1; 230 //cout << l << " " << r << " " << ans << endl; 231 } 232 if (ans == INF) 233 printf ("-1\n"); 234 else 235 printf ("%lld\n", addCost * ans); 236 237 return 0; 238 } 239 */