最短路三巨头

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题目链接：点击打开链接

解题思路：此为最短路径问题

最短时间受人数和路程两个因素影响，因此循环求最短路径时要更新边集。

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;struct aa{int u,v,ub,ug,si1,si2,si3;};void bellman_ford(int n,int k,int D[][31],int i,aa a[],int W[])//知道W【】{/*int D[31],Father[31];for(int i=1;i<=n;i++)D[i]=99999999;D[s]=0;*/for(int node=1;node<=n-1;node++){for(int j=1;j<=2*k;j++){if(D[i][a[j].v]>D[i][a[j].u]+W[j]){D[i][a[j].v]=D[i][a[j].u]+W[j];/*cout<<W[j]<<":"<<j<<endl;cout<<a[j].v<<":"<<a[j].u<<endl;cout<<D[i][a[j].v]<<endl;*/}}}                                            }int main(){int n,b1,b2,b3,g1,g2,g3;cin>>n>>b1>>b2>>b3>>g1>>g2>>g3;int B[31];for(int i=1;i<=n;i++)cin>>B[i];    int G[31];for(int i=1;i<=n;i++)cin>>G[i];int k;cin>>k;aa a[110];for(int i=1;i<=k;i++){cin>>a[i].u>>a[i].v>>a[i].si1>>a[i].si2>>a[i].si3;//a[i].ub=B[a[i].u];//a[i].ug=G[a[i].u];a[i+k].v=a[i].u;a[i+k].u=a[i].v;a[i+k].si1=a[i].si1;    a[i+k].si2=a[i].si3;a[i+k].si3=a[i].si2;//a[i+k].ub=B[a[i].v];//a[i+k].ug=G[a[i].v];}/*int G[31][31];for(int i=1;i<=k;i++)                                        for(int j=1;j<=k;j++)G[i][j]=-1;for(int i=1;i<=k;i++)G[a[i].u][a[i].v]=a[i].si1/b1*a[i].ub+a[i].si1/g1*a[i].ug                  +a[i].si2/b1*a[i].ub+a[i].si2/g1*a[i].ug +a[i].si3/b1*a[i].ub+a[i].si3/g1*a[i].ug;*//*int W[100];//边的权值for(int i=1;i<=2*k;i++){W[i]=a[i].si1*b1*a[i].ub+a[i].si1*g1*a[i].ug    +a[i].si2*b2*a[i].ub+a[i].si2*g2*a[i].ug+a[i].si3*b3*a[i].ub+a[i].si3*g3*a[i].ug;cout<<W[i]<<endl;}*/int D[31][31];for(int i=1;i<=n;i++){int W[110];for(int e=1;e<=2*k;e++){    W[e]=a[e].si1*b1*B[i]+a[e].si1*g1*G[i]+a[e].si2*b2*B[i]+a[e].si2*g2*G[i]+a[e].si3*b3*B[i]+a[e].si3*g3*G[i];//cout<<W[e]<<' ';}//cout<<endl;for(int j=1;j<=n;j++)D[i][j]=99999999;D[i][i]=0;bellman_ford(n,k,D,i,a,W);/*for(int j=1;j<=n;j++)cout<<D[i][j]<<' ';cout<<endl;*/}int min=99999999,ans;for(int i=1;i<=n;i++){int sum=0;for(int j=1;j<=n;j++){sum+=D[j][i];//cout<<sum<<' ';}//cout<<sum<<' ';//cout<<endl;if(sum<min){min=sum;ans=i;}}/*int min=99999999,ans;for(int i=1;i<=n;i++){int D[31];    for(int i2=1;i2<=n;i2++)D[i2]=99999999;D[i]=0;bellman_ford(n,k,D,a,W);int sum=0;for(int j=1;j<=n;j++)sum+=D[j];cout<<sum;if(sum<min){min=sum;ans=a[i].v;}}*/cout<<ans;//system("pause");return 0;}

#include <iostream>using namespace std;struct aa{    int u,v,flowd,up,down,w;};void dijkstra(int s,int k,int n,int D[][31],aa a[]){    for(int i=1;i<=n;i++){            int min=0x7FFFFFFF,u;        for(int i1=1;i1<=n;i1++){            if(D[s][i1]<min){                min=D[s][i1];                u=i1;            }        }        /*从点出发，G[][]式存储        for(int i2=1;i2<=n;i2++){            if        }*/        for(int j=1;j<=2*k;j++){            if(a[j].u==u&&D[s][a[j].v]>D[s][a[j].u]+a[j].w)            D[s][a[j].v]=D[s][a[j].u]+a[j].w;        }    }}int main(){    int n,B1,B2,B3,G1,G2,G3;    cin>>n>>B1>>B2>>B3>>G1>>G2>>G3;    int X[31];    int i;    for(i=1;i<=n;i++)        cin>>X[i];    int Y[31];    for(i=1;i<=n;i++)        cin>>Y[i];    int k;    cin>>k;    aa a[31];    for(i=1;i<=k;i++){        cin>>a[i].u>>a[i].v>>a[i].flowd>>a[i].up>>a[i].down;        a[i+k].u=a[i].v;        a[i+k].v=a[i].u;        a[i+k].flowd=a[i].flowd;        a[i+k].up=a[i].down;        a[i+k].down=a[i].up;    }    int D[31][31];    for(i=1;i<=n;i++){        for(int j=1;j<=2*k;j++){            a[j].w=X[i]*(a[j].flowd*B1+a[j].up*B2+a[j].down*B3)                  +Y[i]*(a[j].flowd*G1+a[j].up*G2+a[j].down*G3);        }        for(int node=1;node<=n;node++)        D[i][node]=0x7FFFFFFF;        D[i][i]=0;        dijkstra(i,k,n,D,a);    }    int min1=0x7FFFFFFF,ans;    for(i=1;i<=n;i++){            int sum=0;        for(int j=1;j<=n;j++){            sum+=D[j][i];        }        if(sum<min1){        min1=sum;        ans=i;        }    }    cout<<ans;    //system("pause");    //getchar();    return 0;}

Floyd要用邻接矩阵存储，不适合此题的数据存储。

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