CodeForcesGym 100753A A Journey to Greece

A Journey to Greece

Time Limit: 5000ms

Memory Limit: 262144KB

This problem will be judged on CodeForcesGym. Original ID: 100753A
64-bit integer IO format: %I64d      Java class name: (Any)

 

解题：状压dp

 #include <bits/stdc++.h>
 using namespace std;
 typedef pair<int,int> pii;
 const int maxn = ;
 struct arc {
     int to,cost,next;
     arc(int x = ,int y = ,int z = -) {
         to = x;
         cost = y;
         next = z;
     }
 } e[maxn*];
 int head[maxn],hs[maxn],tot,N,P,M,G,T;
 void add(int u,int v,int cost) {
     e[tot] = arc(v,cost,head[u]);
     head[u] = tot++;
 }
 priority_queue<pii,vector< pii >,greater< pii > >q;
 int d[][maxn];
 bool done[maxn];
 void dijkstra(int S) {
     while(!q.empty()) q.pop();
     int s = hs[S];
     memset(d[s],0x3f,sizeof d[s]);
     q.push(pii(d[s][S] = ,S));
     memset(done,false,sizeof done);
     while(!q.empty()) {
         int u = q.top().second;
         q.pop();
         if(done[u]) continue;
         done[u] = true;
         for(int i = head[u]; ~i; i = e[i].next) {
             if(d[s][e[i].to] > d[s][u] + e[i].cost) {
                 d[s][e[i].to] = d[s][u] + e[i].cost;
                 q.push(pii(d[s][e[i].to],e[i].to));
             }
         }
     }
 }
 int city[],ttime[],dp[<<][][];
 int main() {
     int u,v,w;
     while(~scanf("%d%d%d%d%d",&N,&P,&M,&G,&T)) {
         memset(head,-,sizeof head);
         bool zero = false;
         for(int i = tot = ; i < P; ++i) {
             scanf("%d%d",city + i,ttime + i);
             hs[city[i]] = i;
             if(city[i] == ) zero = true;
         }
         for(int i = ; i < M; ++i) {
             scanf("%d%d%d",&u,&v,&w);
             add(u,v,w);
             add(v,u,w);
         }
         for(int i = ; i < P; ++i) dijkstra(city[i]);
         if(!zero){
             hs[] = P;
             dijkstra();
         }
         memset(dp,0x3f,sizeof dp);
         for(int i = ; i < P; ++i){
             dp[<<i][i][] = d[hs[]][city[i]] + ttime[i];
             dp[<<i][i][] = T + ttime[i];
         }
         int st = (<<P);
         for(int i = ; i < st; ++i){
             for(int j = ; j < P; ++j){
                 if(((i>>j)&) == ) continue;
                 for(int k = ; k < P; ++k){
                     if((i>>k)&) continue;
                     dp[i|(<<k)][k][] = min(dp[i|(<<k)][k][],dp[i][j][] + d[j][city[k]] + ttime[k]);
                     dp[i|(<<k)][k][] = min(dp[i|(<<k)][k][],dp[i][j][] + T + ttime[k]);
                     dp[i|(<<k)][k][] = min(dp[i|(<<k)][k][],dp[i][j][] + d[j][city[k]] + ttime[k]);
                 }
             }
         }
         bool wtx = false,poss = false;
         for(int i = ; i < P && !wtx; ++i){
             if(dp[st-][i][] + d[i][] <= G) wtx = true;
             if(dp[st-][i][] + T <= G) poss = true;
             if(dp[st-][i][] + d[i][] <= G) poss = true;
         }
         if(wtx) puts("possible without taxi");
         else if(poss) puts("possible with taxi");
         else puts("impossible");
     }
     return ;
 }