POJ 2263 Heavy Cargo(Dijkstra算法)

Heavy Cargo

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1787 Accepted: 1012

Description

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 will contain one or more test cases. The first line of each test case will contain two integers: the number of cities n (2<=n<=200) and the number of road segments r (1<=r<=19900) 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 3Karlsruhe Stuttgart 100Stuttgart Ulm 80Ulm Muenchen 120Karlsruhe Muenchen5 5Karlsruhe Stuttgart 100Stuttgart Ulm 80Ulm Muenchen 120Karlsruhe Hamburg 220Hamburg Muenchen 170Muenchen Karlsruhe0 0

Sample Output

Scenario #180 tons Scenario #2170 tons

Source

Ulm Local 1998

//Dijkstra算法

//每次要找出最大装载的车道，然后连向其他的city[j]，这个最大装载比到city【j】后面的装载比行比较，得出小的装载，再和原来可以直接到达的city[j]的

//装载进行比较，得出较大就是city[j]的最大最终装载

#pragma warning(disable:4768)
#include<iostream>
#include<map>
#include<string>
using namespace std;

#define N 202

int g[N][N], dist[N];
bool visited[N];
map<string, int>m;

int min(int a, int b) {
    return a<b? a:b;
}

int max(int a, int b) {
    return a>b? a:b;
}
void dijkstra(int v, int n)
{
    int i, j;
    for(i = 0; i <= n; i++) {
        dist[i] = g[v][i];
        visited[i] = false;
    }
    visited[v] = true; dist[v] = 0;
    for(i = 1; i < n; i++) {
        int tmp = 0;
        int k = v;
        for(j = 1; j <= n; j++) {
            if((!visited[j]) && (tmp < dist[j])) {
                tmp = dist[j];
                k = j;
            }
        }
        visited[k] = true;
        for(j = 1; j <= n; j++) {
            if((!visited[j]) && (g[k][j] > -1)) {
                dist[j] = max(min(dist[k], g[k][j]), dist[j]);
            }
        }
    }
}    

int main()
{
    int i, j, n, r, x, cnt, C= 0;
    char s1[33], s2[33];
    while(true) {
        cin>>n>>r;
        if(n == 0 && r == 0) break;
        for(i = 0; i <= n; i++)
            for(j = 0; j <= n; j++)
                g[i][j] = -1;
        m.clear();
        for(i = 0, cnt = 1; i < r; i++) {
            cin>>s1>>s2>>x;
            if(!m[s1]) m[s1] = (cnt++);
            if(!m[s2]) m[s2] = (cnt++);
            g[m[s1]][m[s2]] = x;
            g[m[s2]][m[s1]] = x;
        }
        cin>>s1>>s2;
        dijkstra(m[s1], n);
        printf("Scenario #%d\n", ++C);
        printf("%d tons\n", dist[m[s2]]);
        printf("\n");
    }
    return 0;
}