POJ 3268

来源：互联网发布：订房软件编辑：程序博客网时间：2024/06/05 10:46
最短路，使用两次Dijkstra算法，只是注意回来的时候起点和终点就好。
#include<stdio.h>#include<string.h>#include<algorithm>#define INF 1000000using namespace std;int mingo[1005];int minback[1005];int cost[1005][1005];int judge[1005];int n,m,target;void dijgo(int u){    for(int i=1;i<=n;i++) mingo[i]=cost[i][u];    judge[u]=1;   for(int num=1;num<n;num++){        int k=-1;        int min=INF;        for(int i=1;i<=n;i++){            if(judge[i]==0&&min>mingo[i]){                min=mingo[i];                k=i;            }            //if(judge[i]==0) v=i;        }        if(k==-1) break;        judge[k]=1;        for(int i=1;i<=n;i++){            if(judge[i]==0&&mingo[i]>mingo[k]+cost[i][k]){                mingo[i]=mingo[k]+cost[i][k];            }        }        //printf("yes\n");    }    return ;}void dijback(int u){    for(int i=1;i<=n;i++) minback[i]=cost[u][i];    judge[u]=1;    for(int num=1;num<n;num++){        int k=-1;        int min=INF;        for(int i=1;i<=n;i++){            if(judge[i]==0&&min>minback[i]){                min=minback[i];                k=i;            }        }        //if(k==-1) break;        judge[k]=1;        for(int i=1;i<=n;i++){            if(judge[i]==0&&minback[i]>minback[k]+cost[k][i]){                minback[i]=minback[k]+cost[k][i];            }        }        //printf("%d %d\n",k,minback[k]);    }    return ;}int main(){    int x,y,v;    while(scanf("%d%d%d",&n,&m,&target)!=EOF){        for(int i=1;i<=n;i++){            mingo[i]=INF;            minback[i]=INF;            for(int j=1;j<=n;j++){                if(i==j) cost[i][j]=0;                else                    cost[i][j]=INF;            }        }        for(int i=0;i<m;i++){            scanf("%d%d%d",&x,&y,&v);            cost[x][y]=v;        }        //printf("%d %d %d %d\n",cost[4][2],cost[2][1],cost[1][3],cost[3][4]);        memset(judge,0,sizeof(judge));        dijgo(target);        memset(judge,0,sizeof(judge));        dijback(target);        int max=-1;        for(int i=1;i<=n;i++){            //printf("%d %d\n",mingo[i],minback[i]);            if(max<mingo[i]+minback[i]&&i!=target) max=mingo[i]+minback[i];        }        printf("%d\n",max);    }    return 0;}
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