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

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int,int> pii;
 4 const int maxn = 20010;
 5 struct arc {
 6     int to,cost,next;
 7     arc(int x = 0,int y = 0,int z = -1) {
 8         to = x;
 9         cost = y;
10         next = z;
11     }
12 } e[maxn*50];
13 int head[maxn],hs[maxn],tot,N,P,M,G,T;
14 void add(int u,int v,int cost) {
15     e[tot] = arc(v,cost,head[u]);
16     head[u] = tot++;
17 }
18 priority_queue<pii,vector< pii >,greater< pii > >q;
19 int d[20][maxn];
20 bool done[maxn];
21 void dijkstra(int S) {
22     while(!q.empty()) q.pop();
23     int s = hs[S];
24     memset(d[s],0x3f,sizeof d[s]);
25     q.push(pii(d[s][S] = 0,S));
26     memset(done,false,sizeof done);
27     while(!q.empty()) {
28         int u = q.top().second;
29         q.pop();
30         if(done[u]) continue;
31         done[u] = true;
32         for(int i = head[u]; ~i; i = e[i].next) {
33             if(d[s][e[i].to] > d[s][u] + e[i].cost) {
34                 d[s][e[i].to] = d[s][u] + e[i].cost;
35                 q.push(pii(d[s][e[i].to],e[i].to));
36             }
37         }
38     }
39 }
40 int city[20],ttime[20],dp[1<<20][20][2];
41 int main() {
42     int u,v,w;
43     while(~scanf("%d%d%d%d%d",&N,&P,&M,&G,&T)) {
44         memset(head,-1,sizeof head);
45         bool zero = false;
46         for(int i = tot = 0; i < P; ++i) {
47             scanf("%d%d",city + i,ttime + i);
48             hs[city[i]] = i;
49             if(city[i] == 0) zero = true;
50         }
51         for(int i = 0; i < M; ++i) {
52             scanf("%d%d%d",&u,&v,&w);
53             add(u,v,w);
54             add(v,u,w);
55         }
56         for(int i = 0; i < P; ++i) dijkstra(city[i]);
57         if(!zero){
58             hs[0] = P;
59             dijkstra(0);
60         }
61         memset(dp,0x3f,sizeof dp);
62         for(int i = 0; i < P; ++i){
63             dp[1<<i][i][1] = d[hs[0]][city[i]] + ttime[i];
64             dp[1<<i][i][0] = T + ttime[i];
65         }
66         int st = (1<<P);
67         for(int i = 1; i < st; ++i){
68             for(int j = 0; j < P; ++j){
69                 if(((i>>j)&1) == 0) continue;
70                 for(int k = 0; k < P; ++k){
71                     if((i>>k)&1) continue;
72                     dp[i|(1<<k)][k][1] = min(dp[i|(1<<k)][k][1],dp[i][j][1] + d[j][city[k]] + ttime[k]);
73                     dp[i|(1<<k)][k][0] = min(dp[i|(1<<k)][k][0],dp[i][j][1] + T + ttime[k]);
74                     dp[i|(1<<k)][k][0] = min(dp[i|(1<<k)][k][0],dp[i][j][0] + d[j][city[k]] + ttime[k]);
75                 }
76             }
77         }
78         bool wtx = false,poss = false;
79         for(int i = 0; i < P && !wtx; ++i){
80             if(dp[st-1][i][1] + d[i][0] <= G) wtx = true;
81             if(dp[st-1][i][1] + T <= G) poss = true;
82             if(dp[st-1][i][0] + d[i][0] <= G) poss = true;
83         }
84         if(wtx) puts("possible without taxi");
85         else if(poss) puts("possible with taxi");
86         else puts("impossible");
87     }
88     return 0;
89 }