老BOJ 12 Heavy Cargo

最新推荐文章于 2016-04-15 14:48:04 发布

Bupt_Luke

最新推荐文章于 2016-04-15 14:48:04 发布

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分类专栏： BOJ题解老BOJ题解文章标签： dijkstra 老BOJ

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本文介绍了一种使用迪杰斯特拉算法解决寻找两点间最大载重路径的问题，该问题涉及在给定城市间的道路及其重量限制条件下，找到从起点到终点的最大载重能力。

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Heavy Cargo

Accept:85	Submit:228
Time Limit:1000MS	Memory Limit:65536KB

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 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

Floyd or Dijkstra遍历无向图，求出所有连通分量中权重最小的一条边

#include <iostream>
using namespace std;
char name[201][50];
int w[201][201],dis[201];
int sum,n,r;
int getnum(char t[])
{
    for(int i=0; i<=sum; i++)
        if(!strcmp(name[i], t))
            return i;
    sum++;
    strcpy(name[sum], t);
    return sum;
}
int Dijkstra(int s, int e)
{
    int i,j,t,minw;
    bool flag[201]= {0};
    for(i=0; i<n; i++)
        dis[i] = w[s][i];
    dis[s]=0;
    flag[s]=true;
    for(i=0; i<n; i++)
    {
        minw = dis[e];
        t = s;
        for(j=0; j<n; j++)
            if(dis[j]>minw && !flag[j])
            {
                t = j;
                minw = dis[j];
            }
        if(flag[t]) return dis[e];
        flag[t] = true;
        for(j=0; j<n; j++)
            if(!flag[j] && w[t][j]>-1)
            {
                minw = dis[t]>w[t][j]?w[t][j]:dis[t];
                dis[j] = dis[j]<minw?minw:dis[j];
            }
    }
    return dis[e];
}
int main()
{
    int i,weight,s,e,Case=0;
    char a[50],b[50];
    while(scanf("%d %d%*c", &n, &r),n+r)
    {
        memset(w,-1,sizeof(w));
        memset(name,'\0',sizeof(name));
        sum=-1;
        for(i=0; i<r; i++)
        {
            scanf("%s %s %d", &a, &b, &weight);
            s = getnum(a);
            e = getnum(b);
            w[s][e]=w[e][s]=weight;
        }
        scanf("%s %s", &a, &b);
        s = getnum(a);
        e = getnum(b);
        printf("Scenario #%d\n%d tons\n\n",++Case, Dijkstra(s,e));
    }
    return 0;
}