| Heavy Cargo |
Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount of cargo you can transport with it is never limited by the truck itself. It is only limited by the weight restrictions that apply for the roads along the path you want to drive.
Given start and destination city, your job is to determine the maximum load of the Godzilla V12 so
that there still exists a path between the two specified cities.
Input
The input file will contain one or more test cases. The first line of each test case will contain two
integers: the number of cities n (
)
and the number of road segments r (
)
making up the street network.
Then r lines will follow, each one describing one road segment by naming the two cities connected by the segment and giving the weight limit for trucks that use this segment. Names are not longer than 30 characters and do not contain white-space characters. Weight limits are integers in the range 0 - 10000. Roads can always be travelled in both directions.
The last line of the test case contains two city names: start and destination.
Input will be terminated by two values of 0 for n and r.
Output
For each test case, print three lines:
- a line saying ``Scenario #x" where x is the number of the test case
- a line saying ``y tons" where y is the maximum possible load
- a blank line
Sample Input
4 3 Karlsruhe Stuttgart 100 Stuttgart Ulm 80 Ulm Muenchen 120 Karlsruhe Muenchen 5 5 Karlsruhe Stuttgart 100 Stuttgart Ulm 80 Ulm Muenchen 120 Karlsruhe Hamburg 220 Hamburg Muenchen 170 Muenchen Karlsruhe 0 0
Sample Output
Scenario #1 80 tons Scenario #2 170 tons
Miguel A. Revilla
1999-01-11 dijkstra变形，应该也可以n^2算法dfs。另外，最大流貌似也可做。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
using namespace std;
#define LL long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define cint const int
#define INF 100000000
#define MAXN 220
#define MAXM 40000
struct edge{
int u, v, w, nxt;
}e[MAXM];
int h[MAXN], cc, n, m;
map<string, int> mp;
int cnt;
int U(string str){
if(mp.count(str)) return mp[str];
else return mp[str] = cnt++;
}
void add(int u, int v, int w){
e[cc]=(edge){u, v, w, h[u]};
h[u]=cc++;
e[cc]=(edge){v, u, w, h[v]};
h[v]=cc++;
}
struct node{
int u, d;
bool operator < (const node &rhs)const{
return d < rhs.d;
}
};
int d[MAXN];
bool done[MAXN];
int Dijkstra(cint s, cint t){
priority_queue<node> q;
fill_n(d, n, 0); d[s]=INF;
fill_n(done, n, false); q.push((node){s, d[s]});
while(!q.empty()){
node ut = q.top(); q.pop();
int u = ut.u; if(t==u) return d[t];
if(done[u]) continue; done[u]=true;
for(int i=h[u]; i!=-1; i=e[i].nxt){
int v = e[i].v, w = e[i].w;
d[v] = max(d[v], min(d[u], w));
q.push((node){v, d[v]});
}
}
return d[t];
}
int main(){
// freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
int kase=1;
while(scanf(" %d %d", &n, &m)==2 && (n || m)){
int i, w;
fill_n(h, n, -1); cc=0;
mp.clear(); cnt=0;
char s1[33], s2[33];
for(i=0; i<m; i++){
scanf(" %s %s %d", s1, s2, &w);
add(U(string(s1)), U(string(s2)), w);
}
scanf(" %s %s", s1, s2);
printf("Scenario #%d
%d tons
", kase++, Dijkstra(U(s1), U(s2)));
}
return 0;
}