poj [1753]

棋盘的上的每一个状态对应一个二进制串，以初始状态为跟节点建树，然后广度优先遍历这颗树，层数为所求的值。

#include<iostream>   
#include<queue>   
#define ALL_WHITE 0   
#define ALL_BLACK 65535   
using namespace std;   
int states[65536];   
int state,temp;   
int flip(int state_current, int pos) {   
    state_current ^= (1 << pos);   
    if (pos - 4 >= 0) {   
        state_current ^= (1 << (pos - 4));   
    }   
    if (pos + 4 < 16) {   
        state_current ^= (1 << (pos + 4));   
    }   
    if (pos % 4 != 0) {   
        state_current ^= (1 << (pos - 1));   
    }   
    if (pos % 4 != 3) {   
        state_current ^= (1 << (pos + 1));   
    }   
    return state_current;   
}   
void bfs() {   
    int i;   
    memset(states,-1,sizeof(states));   
    queue<int> q;   
    states[state] = 0;   
    q.push(state);   
    if (state == ALL_BLACK || state == ALL_WHITE){   
        cout<<"0"<<endl;   
        return ;   
    }   
    while(!q.empty()){   
        temp=q.front();   
        q.pop();   
        for(i=0;i<16;i++) {   
            int next_state = flip(temp, i);   
            if (next_state == ALL_BLACK || next_state == ALL_WHITE){   
                cout<<states[temp]+1<<endl;   
                return ;   
            }   
            if (states[next_state] == -1){   
                states[next_state] = states[temp] + 1;   
                q.push(next_state);   
            }   
        }   
    }   
    cout<<"Impossible"<<endl;   
}   
int main(){   
    char c;   
    int i;   
    state = 0;   
    for(i=0; i<16; i++){   
        cin>>c;   
        if(c=='b'){   
            state += 1<<i;   
        }   
        if (c == 'w'){   
            state += 0<<i;   
        }   
    }   
    bfs();   
    return 0;   
}