HDU 1385Minimum Transport Cost 最短路输出路径

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1385

题意

给定n个城镇，从s到t运货需要路费，路径中除了s和t之外，经过每个城镇还要额外收税。求s到t的最小花费，并输出字典序最小的路径

思路

一开始从s到t正向跑dj，发现倒着走反而才是字典序最小的，要输出从s->t的最短路且字典序最小，从t倒着走到s，每次拿编号最小的点去松弛其他点。因为s和t不计算税收，所以需先将s的税暂时赋0
dj版

#include<cstdio>#include<queue>#include<iostream>#include<vector>#include<map>#include<cstring>#include<string>#include<set>#include<stack>#include<algorithm>#define cle(a) memset(a,0,sizeof(a))#define inf(a) memset(a,0x3f,sizeof(a))#define ll long long#define Rep(i,a,n) for(int i=a;i<=n;i++)using namespace std;#define INF2 9223372036854775807llconst int INF = ( 2e9 ) + 2;const ll maxn = 1e4+10;int mp[maxn][maxn];int p[maxn],d[maxn],vis[maxn],pre[maxn];int n;void dijkstra(int s,int t){    for(int i=1; i<=n; i++)d[i]=INF;    d[t]=0;    memset(vis,0,sizeof(vis));    memset(pre,-1,sizeof(pre));    for(int i=1; i<=n; i++)    {        int u=-1,mn=INF;        for(int j=n; j>=1; j--)            if(!vis[j]&&d[j]<mn)                mn=d[u=j];        if(u==-1)break;        vis[u]=1;        for(int v=n; v>=1; v--)            if(mp[v][u]!=-1&&u!=v)            {                if(d[v]>d[u]+mp[v][u]+p[v])                {                    d[v]=d[u]+mp[v][u]+p[v];                    pre[v]=u;                }                else if(d[v]==d[u]+mp[v][u]+p[v])                {                    pre[v]=min(pre[v],u);                }            }    }    printf("From %d to %d :\n",s,t);    printf("Path: ");    for(int i=s; i!=-1; i=pre[i])        if(i==s)            printf("%d",i);        else            printf("-->%d",i);    printf("\nTotal cost : %d\n",d[s]);}int main(){//  freopen("in.txt","r",stdin);    while(~scanf("%d",&n)&&n)    {        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)                scanf("%d",&mp[i][j]);        for(int i=1; i<=n; i++)            scanf("%d",&p[i]);        int u,v;        u=v=0;        while(1)        {            scanf("%d%d",&u,&v);            if(u==-1&&v==-1)break;            int temp=p[u];            p[u]=0;            dijkstra(u,v);            printf("\n");            p[u]=temp;        }    }}

Floyd版

#include<cstdio>#include<queue>#include<iostream>#include<vector>#include<map>#include<cstring>#include<string>#include<set>#include<stack>#include<algorithm>#define cle(a) memset(a,0,sizeof(a))#define inf(a) memset(a,0x3f,sizeof(a))#define ll long long#define Rep(i,a,n) for(int i=a;i<=n;i++)using namespace std;#define INF2 9223372036854775807llconst int INF = ( 2e9 ) + 2;const ll maxn = 110;int mp[maxn][maxn],path[maxn][maxn],p[maxn];int cnt=0;void printpath(int u,int v){    if(u==v)    {        printf("%d\n",v);        return;    }    else    {        int k=path[u][v];        printf("%d-->",u);        printpath(k,v);    }}int main(){    int n;//  freopen("in.txt","r",stdin);    while(~scanf("%d",&n)&n)    {        for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        {            scanf("%d",&mp[i][j]);            path[i][j]=j;            if(mp[i][j]==-1)            mp[i][j]=INF;        }        for(int i=1;i<=n;i++)        scanf("%d",&p[i]);        for(int k=1;k<=n;k++)        for(int i=1;i<=n;i++)        {            if(mp[i][k]==INF)continue;            for(int j=1;j<=n;j++)            if(mp[j][k]!=INF)            {                if(mp[i][j]>mp[i][k]+mp[k][j]+p[k])                {                    mp[i][j]=mp[i][k]+mp[k][j]+p[k];                    path[i][j]=path[i][k];                }                else if(mp[i][j]==mp[i][k]+mp[k][j]+p[k])                {                    if(path[i][j]>path[i][k])                    path[i][j]=path[i][k];                }            }        }        int u,v;        while(1)        {            scanf("%d%d",&u,&v);            if(u==-1&&v==-1)break;            printf("From %d to %d :\n",u,v);            printf("Path: ");            printpath(u,v);            printf("Total cost : %d\n\n",mp[u][v]);        }    }}