M67的题目：情人迷宫

原题 http://www.matrix67.com/blog/archives/550一个广搜就过了。合法解：设mm从A出发，gg从B出发 mm->I gg->A mm->H gg->I mm->G gg->H mm->F->L gg->I mm->E->D gg->H->J mm->C gg->K mm->B gg->B

#include <cstdlib>#include <iostream>#include <fstream>using namespace std;const int MAXINT = 1 << 30;int n;char vc[12]; //点的颜色 int e[12]; int p[12][3]; char c[12][3]; //邻接表 const int QSIZE = 100000;int q[QSIZE][5], qf, qr, qn;int steps[12][12]; //最优化 inline void enqueue(int a, int b, int s, int prva, int prvb){    q[qr][0] = a; q[qr][1] = b; q[qr][2] = s;    q[qr][3] = prva; q[qr][4] = prvb;     qr++; qr %= QSIZE;    qn++;}void dequeue(int &a, int &b, int &s, int &prva, int &prvb){    a = q[qf][0]; b = q[qf][1]; s = q[qf][2];    prva = q[qf][3]; prvb = q[qf][4];    qf++; qf %= QSIZE;    qn--;}int main(){    ifstream fin("valentine.txt");    fin >> n;    for (int i=0; i<n; i++) fin >> vc[i]; //点的颜色，C(yan) R(ed) Y(ellow)     int ecount;    fin >> ecount;    for (int i=0; i<n; i++) e[i] = 0;    for (int i=0; i<ecount; i++) //边集输入，格式为"字母1 字母2 颜色"    {        char u, v, tc;        fin >> u >> v >> tc;        u -= 'A';        v -= 'A';        p[u][e[u]] = v;        p[v][e[v]] = u;        c[u][e[u]] = c[v][e[v]] = tc;        e[u]++; e[v]++;    }    fin.close();        for (int i=0; i<n; i++)        for (int j=0; j<n; j++)            steps[i][j] = MAXINT;        qf = qr = qn = 0;    enqueue(0, 1, 0, -1, -1); //start from Va and Vb    int a, b, s, prva, prvb;    while (qn > 0)    {        dequeue(a, b, s, prva, prvb);        if (steps[a][b] > s) //较优移动         {            cout<<char(a+'A')<<" "<<char(b+'A')<<" "<<s                <<"/t("<<char(prva+'A')<<" "<<char(prvb+'A')<<")"<<endl;            steps[a][b] = s;                        if (a == b) //合法解             {                cout<<"FIND SOL AT V"<<a<<" STEP="<<s<<endl;                continue;            }        }        else //较劣移动             continue;                //move a        for (int i=0; i<e[a]; i++)            if (c[a][i] == vc[b])                enqueue(p[a][i], b, s+1, a, b);                //move b        for (int i=0; i<e[b]; i++)            if (c[b][i] == vc[a])                enqueue(a, p[b][i], s+1, a, b);    }    system("pause");    return 0;}/*valentine.txt12RYCRYCRYCYRR15A B CB C RC D YD E CE F RF G YG H CH I RI A YH J RB K YE L CF L YJ K CK L R*/