【Floyd算法】Gym - 101572I - Import Spaghetti

题意：有向图最小环，输出方案。

#include<cstdio>
#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<cstring>
using namespace std;
int n;
map<string,int>ma;
string mb[505];
int a[505][505],ans=2147483647;
int path[505][505];
int main(){
    //freopen("i.in","r",stdin);
    memset(a,0x7f,sizeof(a));
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        cin>>mb[i];
        ma[mb[i]]=i;
    }
    string x,z;
    int y;
    for(int i=1;i<=n;++i){
        cin>>x>>y;
        if(!y){
            continue;
        }
        int ma_x=ma[x];
        int cnt=0;
        while(1){
            cin>>z;
            if(z=="import"){
                continue;
            }
            if(z[z.length()-1]!=','){
                ++cnt;
            }
            else{
                z.erase(z.length()-1,1);
            }
            if(ma[z]==ma_x){
                cout<<z<<endl;
                return 0;
            }
            a[ma_x][ma[z]]=1;
            path[ma_x][ma[z]]=ma[z];
            if(cnt==y){
                break;
            }
        }
    }
    for(int i=1;i<=n;++i){
        for(int j=1;j<=n;++j){
            for(int k=1;k<=n;++k){
                if(i!=j && j!=k && a[j][i]<2000000000 && a[i][k]<2000000000){
                    if(a[j][i]+a[i][k]<a[j][k]){
                        a[j][k]=a[j][i]+a[i][k];
                        path[j][k]=path[j][i];
                    }
                    else if(a[j][i]+a[i][k]==a[j][k]){
                        if(path[j][k]==0){
                            path[j][k]=path[j][i];
                        }
                    }
                }
            }
        }
    }
    int I,J;
    for(int i=1;i<=n;++i){
        for(int j=1;j<=n;++j){
            if(i!=j){
                if(ans-a[i][j]>a[j][i]){
                    ans=a[i][j]+a[j][i];
                    I=i;
                    J=j;
                }
            }
        }
    }
    if(ans>2000000000){
        puts("SHIP IT");
        return 0;
    }
    int U=I;
    cout<<mb[U];
    while(U!=J){
        cout<<' '<<mb[path[U][J]];
        U=path[U][J];
    }
    U=J;
    while(path[U][I]!=I){
        cout<<' '<<mb[path[U][I]];
        U=path[U][I];
    }
    puts("");
    return 0;
}

　　附无向图最小环（转自 http://www.cnblogs.com/kane0526/archive/2012/11/09/2763170.html）：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int maxn=110;
const int INF=0x7ffffff;
int  dist[maxn][maxn], map[maxn][maxn];
int  pre[maxn][maxn];
int  path[maxn];
int  n, m, num, minc;

void floyd()
{
    minc=INF;
    for(int k=1; k<=n; k++)
    {
        for(int i=1; i<k; i++)
            for(int j=i+1; j<k; j++)
            {
                int  ans=dist[i][j]+map[i][k]+map[k][j];
                if(ans<minc)  //找到最优解
                {
                    minc=ans;
                    num=0;
                    int p=j;
                    while(p!=i)  //逆向寻找前驱遍历的路径并将其存储起来
                    {
                        path[num++]=p;
                        p=pre[i][p];
                    }
                    path[num++]=i;
                    path[num++]=k;
                }
            }
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                if(dist[i][j]>dist[i][k]+dist[k][j])
                {
                    dist[i][j]=dist[i][k]+dist[k][j];
                    pre[i][j]=pre[k][j];
                }
            }
    }
}

int main()
{
    int  u, v, cost;
    while(cin >> n)
    {
        if(n<0) break;
        cin >> m;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                dist[i][j]=map[i][j]=INF;
                pre[i][j]=i;
            }
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&u,&v,&cost);
            if(dist[u][v]>cost)   //处理重边
                map[u][v]=map[v][u]=dist[u][v]=dist[v][u]=cost;
        }
        floyd();
        if(minc==INF)
            printf("No solution.\n");
        else
        {
            printf("%d",path[0]);
            for(int i=1; i<num; i++)
                printf(" %d",path[i]);
            puts("");
        }
    }
    return 0;
}