修道士和野人问题

优化版 v1.0

i7 6500U 4G内存 2s内可以对 1000 2000 输入建图

建图复杂度取决于边数。

description：

假设有n个修道士和n个野人准备渡河，但只有一条能容纳c人的小船，为了防止野人侵犯修道士，要求无论在何处，修道士的个数不得少于野人的人数（除非修道士个数为0）。如果两种人都会划船，试设计一个算法，确定他们能否渡过河去，若能，则给出一个小船来回次数最少的最佳方案并求出所有的解。

sample input：

3 2 //野人修道士各三个，船限载2人

4 2

1000 2000

#include <bits/stdc++.h>using namespace std;const int MAXN=1e7+10;typedef struct Node{    int x1,x2;bool dir;    bool operator < (const Node &a)const{        if(x1==a.x1){            if(x2==a.x2){                return dir<a.dir;            }else return x2<a.x2;        }else return x1<a.x1;    }}Node;int n,k,tot;map <Node,int> mp;Node nodes[MAXN];int fa[MAXN],dep[MAXN];vector <int> edge[MAXN];int vis[MAXN];int sta[MAXN],head;bool output_control;void ini(){    memset(dep,0,sizeof(dep));memset(fa,-1,sizeof(fa));memset(vis,0,sizeof(vis));    mp.clear();for(int i=0;i<=tot;i++) edge[i].clear();    tot=0;head=-1;}void build(){    int lim1,lim2;    queue <Node> que;Node now={n,n,0},pos;    que.push(now);    nodes[++tot]=now;    mp[now]=tot;    fa[tot]=-1;    bool lim_control=0;    while(!que.empty()){        now=que.front();que.pop();        int x=mp[now],y;        lim1=now.x1-now.x2;        lim2=(n-now.x1)-(n-now.x2);        if(now.dir){            pos.dir=0;            if(now.x1!=0) lim_control=1;            else lim_control=0;            for(int j=1;j<=n-now.x2&&j<=k;j++){//i==0                if(lim_control&&j>lim1) break;                pos.x1=now.x1;pos.x2=now.x2+j;                if(mp[pos]==0){                    mp[pos]=++tot;                    nodes[tot]=pos;                    if(pos.x1!=0||pos.x2!=0) que.push(pos);                }                y=mp[pos];                edge[x].push_back(y);            }            for(int i=1;i<n-now.x1&&i<=k;i++){//i!=0                for(int j=max(0,i-lim2);j<=i&&j<=n-now.x2&&(i+j)<=k&&(j-i)<=lim1;j++){                    pos.x1=now.x1+i;pos.x2=now.x2+j;                    if(mp[pos]==0){                        mp[pos]=++tot;                        nodes[tot]=pos;                        if(pos.x1!=0||pos.x2!=0) que.push(pos);                    }                    y=mp[pos];                    edge[x].push_back(y);                }            }            if(now.x1!=n)            for(int i=n-now.x1;i<=n-now.x1&&i<=k;i++){                for(int j=0;j<=i&&j<=n-now.x2&&(i+j)<=k&&(j-i)<=lim1;j++){                    pos.x1=now.x1+i;pos.x2=now.x2+j;                    if(mp[pos]==0){                        mp[pos]=++tot;                        nodes[tot]=pos;                        if(pos.x1!=0||pos.x2!=0) que.push(pos);                    }                    y=mp[pos];                    edge[x].push_back(y);                }            }        }else{            pos.dir=1;            if(now.x1!=n) lim_control=1;            else lim_control=0;            for(int j=1;j<=now.x2&&j<=k;j++){//i==0                if(lim_control&&j>lim2) break;                pos.x1=now.x1;pos.x2=now.x2-j;                if(mp[pos]==0){                    mp[pos]=++tot;                    nodes[tot]=pos;                    if(pos.x1!=0||pos.x2!=0) que.push(pos);                }                y=mp[pos];                edge[x].push_back(y);            }            for(int i=1;i<now.x1&&i<=k;i++){//i<now.x1                for(int j=max(0,i-lim1);j<=i&&j<=now.x2&&(i+j)<=k&&(j-i)<=lim2;j++){                    pos.x1=now.x1-i;pos.x2=now.x2-j;                    if(mp[pos]==0){                        mp[pos]=++tot;                        nodes[tot]=pos;                        if(pos.x1!=0||pos.x2!=0) que.push(pos);                    }                    y=mp[pos];                    edge[x].push_back(y);                }            }            if(now.x1!=0)            for(int i=now.x1;i<=now.x1&&i<=k;i++){//i<now.x1                for(int j=0;j<=i&&j<=now.x2&&(i+j)<=k&&(j-i)<=lim2;j++){                    pos.x1=now.x1-i;pos.x2=now.x2-j;                    if(mp[pos]==0){                        mp[pos]=++tot;                        nodes[tot]=pos;                        if(pos.x1!=0||pos.x2!=0) que.push(pos);                    }                    y=mp[pos];                    edge[x].push_back(y);                }            }        }    }}void show_way(int x){    if(x==-1) return ;    show_way(fa[x]);    cout<<"Node:  "<<x<<"  position1:  "<<nodes[x].x1<<"  "<<nodes[x].x2<<"  position2:  "<<n-nodes[x].x1<<"  "<<n-nodes[x].x2<<"  direction:  "<<nodes[x].dir<<endl;}void solve(){    int now,pos,len,lim=tot+1;    queue <int> que;    que.push(1);    dep[1]=1;    while(!que.empty()){        now=que.front();        que.pop();        len=edge[now].size();        for(int i=0;i<len;i++){            pos=edge[now][i];            if(!dep[pos]){                dep[pos]=dep[now]+1;                fa[pos]=now;                if(edge[pos].empty()){                    cout<<"We need at least "<<dep[pos]-1<<" steps to solve the problem."<<endl;                    show_way(pos);                    cout<<endl;                    return ;                }                else que.push(pos);            }        }    }    if(lim==tot+1) puts("Sorry, no answer");}void show_path(){    cout<<"feasible path: "<<endl;    for(int i=0;i<=head;i++)        cout<<"Node:  "<<sta[i]<<"  position1:  "<<nodes[sta[i]].x1<<"  "<<nodes[sta[i]].x2<<"  position2:  "<<n-nodes[sta[i]].x1<<"  "<<n-nodes[sta[i]].x2<<"  direction:  "<<nodes[sta[i]].dir<<endl;    cout<<endl;}int dfs(int now,int left){    vis[now]=1;    sta[++head]=now;    int len=edge[now].size(),pos;    int ans=0;    if(len==0){        if(!output_control) show_path();        ans++;    }else{        if(left>0){            for(int i=0;i<len;i++){                pos=edge[now][i];                if(!vis[pos])                ans+=dfs(pos,left-1);            }        }    }    vis[now]=0;    head--;    return ans;}void search_all(){    cout<<"Please enter the step limit: ";    int p;cin>>p;cout<<endl;    cout<<"Print all the path?(Y/N): ";    char t;cin>>t;    output_control=(t=='Y')?0:1;    head=-1;    cout<<"There are total "<<dfs(1,p)<<" answer(s) with the step limit of "<<p<<"."<<endl;}void debug(){    cout<<"tot: "<<tot<<endl;    int edges=0;    for(int i=1;i<=tot;i++)        edges+=edge[i].size();    cout<<"edges: "<<edges<<endl;    cout<<"show nodes?(Y/N)"<<endl;    char t;cin>>t;    if(t=='Y')        for(int i=1;i<=tot;i++)            cout<<i<<": "<<nodes[i].x1<<' '<<nodes[i].x2<<' '<<nodes[i].dir<<endl;    cout<<"show edges?(Y/N)"<<endl;    cin>>t;    if(t=='Y')        for(int i=1;i<=tot;i++){            cout<<i<<": ";            for(int j=0;j<edge[i].size();j++)                cout<<edge[i][j]<<' ';            cout<<endl;        }}int main(){   while(1){        system("cls");        cout<<"please enter the number of monk(s) or barbarian(s) and load limit of ship ( please separate them by space ):  ";        cin>>n>>k;        cout<<endl;        build();        while(1){            cout<<"1.Find the shortest path"<<endl;            cout<<"2.Find the all the path with step limit x"<<endl;            cout<<"3.Change the input"<<endl;            cout<<"4.Developer Mode"<<endl;            cout<<"Please enter 1 2 3 or 4 to continue: ";            int p;            cin>>p;            if(p==1) solve();            else if(p==2) search_all();            else if(p==4) debug();            else break;            system("pause");        }        ini();    }}

当然我们也可以不建图,直接bfs求最短路或dfs求定长路径。复杂度会减小很多，改改就行。