FZU 2180 双向BFS

来源：互联网发布：农夫山泉瓶子尺寸数据编辑：程序博客网时间：2024/05/22 11:56
原理很简单，见代码
#include <stdio.h>#include <iostream>#include <algorithm>#include <vector>#include <map>#include <set>#include <cstring>#include <deque>#define INF 0x3f3f3f3fusing namespace std;const int dx[] = {-1, -1, 1, 1, 2, 2, -2, -2 };const int dy[] = {-2, 2, 2, -2, 1, -1, 1, -1 };string G[5] = {    "11111",    "01111",    "00*11",    "00001",    "00000"};string dp[5];map<pair<int,int>, int> mp1;map<pair<int,int>, int> mp2;deque<pair<int,int> > d1, d2;bool inside(int i, int j) {    return i<5 && i>=0 && j<5 && j>=0;}int BFS() {    while (1) {        int firstok = true;        int secondok = true;        if (d1.empty()==false) {            //第一个进行bfs            pair<int,int> top1 = d1.front();            d1.pop_front();            if (mp2.find(top1)!=mp2.end()) {                return mp1[top1] + mp2[top1];            }            //更新第一个队列            for (int k=0; k<8; k++) {                int x = top1.second/5;                int y = top1.second%5;                int nx = top1.second/5 + dx[k];                int ny = top1.second%5 + dy[k];                if (inside(nx, ny)) {                    int ans = top1.first;                    if (top1.first & (1<<(nx * 5 + ny))) {                        //相应的位置交换01                        ans = ans | (1<<(x * 5 + y));                        ans = ans & ~ (1<<(nx * 5 + ny));                    } else {                    }                    int pos = nx * 5 + ny;                    if (mp1.find(pair<int,int>(ans, pos))!=mp1.end()) continue;                    int tmp = mp1[top1] + 1;                    if (tmp > 8) continue;                    mp1[pair<int,int>(ans, pos)] = mp1[top1] + 1;                    d1.push_back(pair<int,int>(ans, pos));                }            }        } else {            firstok = false;        }        {            if (d2.empty()==false) {            //第2个进行bfs                pair<int,int> top2 = d2.front();                d2.pop_front();                if (mp1.find(top2)!=mp1.end()) {                    return mp2[top2] + mp1[top2];                }                //更新第一个队列                for (int k=0; k<8; k++) {                    int x = top2.second/5;                    int y = top2.second%5;                    int nx = top2.second/5 + dx[k];                    int ny = top2.second%5 + dy[k];                    if (inside(nx, ny)) {                        int ans = top2.first;                        if (top2.first & (1<<(nx * 5 + ny))) {                            //相应的位置交换01                            ans = ans | (1<<(x * 5 + y));                            ans = ans & ~ (1<<(nx * 5 + ny));                        } else {                        }                        int pos = nx * 5 + ny;                        if (mp2.find(pair<int,int>(ans, pos))!=mp2.end()) continue;                        int tmp = mp2[top2] + 1;                        if (tmp > 8) continue;                        mp2[pair<int,int>(ans, pos)] = mp2[top2] + 1;                        d2.push_back(pair<int,int>(ans, pos));                    }                }            } else {                secondok = false;            }        }        if (firstok == false && secondok == false) break;    }    return -1;}int main() {    //freopen("in.txt", "r", stdin);     int T;     cin >> T;     while (T--) {        int bx, by;        for (int i=0; i<5; i++) {            cin >> dp[i];            for (int j=0; j<5; j++) {                if (dp[i][j] == '*') {                    bx = i, by = j;                }            }        }        mp1.clear(), mp2.clear();        d1.clear(), d2.clear();        int ans = 0;        for (int i=0; i<5; i++) {            for (int j=0; j<5; j++) {                if (G[i][j] == '1') {                    ans = ans | (1<<(i*5+j));                }            }        }        int pos = 2*5+2;        mp1[pair<int, int>(ans, pos)] = 0;        d1.push_back(pair<int,int>(ans, pos));        ans = 0;        for (int i=0; i<5; i++) {            for (int j=0; j<5; j++) {                if (dp[i][j] == '1') {                    ans = ans | (1<<(i*5+j));                }            }        }        pos = bx * 5 + by;        mp2[pair<int,int>(ans, pos)] = 0;        d2.push_back(pair<int,int>(ans, pos));        int ret = BFS();        if (ret == -1 || ret > 15) {            cout << "Bored!" << endl;        } else {            cout << ret << endl;        }     }}
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