标签：acm   算法   最短路   

HDU Today

6
xiasha westlake
xiasha station 60
xiasha ShoppingCenterofHangZhou 30
station westlake 20
ShoppingCenterofHangZhou supermarket 10
xiasha supermarket 50
supermarket westlake 10
-1

50


Hint:
The best route is:
xiasha->ShoppingCenterofHangZhou->supermarket->westlake


虽然偶尔会迷路，但是因为有了你的帮助
**和**从此还是过上了幸福的生活。

――全剧终――

基本上是floyd-warshall的模板题，直接调用函数，注意map的使用即可，还要特别判断n=0的情况。

#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma commment(linker,"/STACK: 102400000 102400000")
#define mset0(t) memset(t,0,sizeof(t))
#define lson a,b,l,mid,cur<<1
#define rson a,b,mid+1,r,cur<<1|1
using namespace std;
const double eps=1e-5;
const int MAXN=165;
const int INF=0x3f3f3f3f;

int d[MAXN][MAXN],prev[MAXN],n,v;


void warshal_floyd()
{
    for(int k=1; k<=v; k++)
        for(int i=1; i<=v; i++)
            for(int j=1; j<=v; j++)
                d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
}

int main()
{
    while(scanf("%d",&n)&&n!=-1)
    {
        map<string,int>m;
        for(int i=1; i<=165; i++)
            for(int j=i; j<=165; j++)
            {
                if(i==j)
                    d[i][j]=0;
                else
                    d[i][j]=d[j][i]=INF;
            }
        string s,e,a,b;
        int cnt=0,t;
        cin>>s>>e;
        if(n==0)//特判有0条路的情况
        {
            if(s!=e)
                printf("-1\n");
            else
                printf("0\n");
            continue;
        }
        m[s]=++cnt;
        if(m[e]==0)
            m[e]=++cnt;
        for(int i=0; i<n; i++)
        {
            cin>>a>>b>>t;
            if(m[a]==0)
                m[a]=++cnt;
            if(m[b]==0)
                m[b]=++cnt;
            d[m[a]][m[b]]=d[m[b]][m[a]]=t;
        }
        v=cnt;
        warshal_floyd();
        printf("%d\n",d[m[s]][m[e]]==INF?-1:d[m[s]][m[e]]);
    }
    return 0;
}

HDU 2112 HDU Today(floyd-warshall+map)

标签：acm   算法   最短路   

原文地址：http://blog.csdn.net/noooooorth/article/details/47051075