[UVA] 544 - Heavy Cargo＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2012-04-16 22:39:42| 人1,590| 回0 | 上一篇 | 下一篇

[UVA] 544 - Heavy Cargo

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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

SPFA


#include <iostream>
#include <stdio.h> 
#include <string.h>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef struct {
 int to,v;
} Arc;
typedef vector<Arc>::iterator it;
vector<Arc> Link[201];

int Min(int x, int y) {
 return x < y ? x : y;
}
int SPFA(int st, int ed, int n) {
 queue<int> Q;
 int dis[201];
 int used[201];
 memset(dis, 0, sizeof(dis));
 memset(used, 0, sizeof(used));
 dis[st] = 0xfffffff;
 for(it i = Link[st].begin(); i != Link[st].end(); i++) {
 if(dis[i->to] < Min(dis[st], i->v)) {
 dis[i->to] = Min(dis[st], i->v);
 if(used[i->to] == 0) {
 used[i->to] = 1;
 Q.push(i->to);
 }
 }
 }
 int tv;
 while(!Q.empty()) {
 tv = Q.front();
 Q.pop();
 used[tv] = 0;
 for(it i = Link[tv].begin(); i != Link[tv].end(); i++) {
 if(dis[i->to] < Min(dis[tv], i->v)) {
 dis[i->to] = Min(dis[tv], i->v);
 if(used[i->to] == 0) {
 used[i->to] = 1;
 Q.push(i->to);
 }
 }
 }
 }
 return dis[ed];
}
int main() {
 int n, r, Case = 0, i;
 while(scanf("%d %d", &n, &r) == 2) {
 if(n == 0 && r == 0)
 break;
 map<string, int> record;
 for(i = 0; i <= n; i++)
 Link[i].clear();
 int encode = 0, v, nx, ny;
 string x, y;
 while(r--) {
 cin >> x >> y >> v;
 if(record[x] == 0) { 
 encode ++;
 record[x] = encode;
 } 
 if(record[y] == 0) {
 encode++;
 record[y] = encode;
 }
 nx = record[x], ny = record[y];
 Arc arc;
 arc.to = ny, arc.v = v;
 Link[nx].push_back(arc);
 arc.to = nx, arc.v = v;
 Link[ny].push_back(arc);
 }
 cin >> x >> y; 
 printf("Scenario #%d\n", ++Case);
 nx = record[x], ny = record[y];
 printf("%d tons\n", SPFA(nx, ny, n));
 puts("");
 }
 return 0;
}