【暴力枚舉&BFS】Flow Free @RMRC2017/upcexam5124

時間限制: 1 Sec 內存限制: 128 MB
題目描述
Flow Free is a puzzle that is played on a 2D grid of cells, with some cells marked as endpoints of certain colors and the rest of cells being blank. To solve the puzzle, you have to connect each pair of colored endpoints with a path, following these rules:
there is a path connecting two points with the same color, and that path also has that color
all cells in the grid are used and each cell belongs to exactly one path (of the same color as the endpoints of the path)
The rules imply that different colored paths cannot intersect.
The path is defined as a connected series of segments where each segment connects two neighbouring cells. Two cells are neighbours if they share a side (so all segments are either horizontal or vertical). By these definitions and rules above, each colored cell will be an endpoint of exactly one segment and each blank cell will be an endpoint of exactly two segments.
In this problem we will consider only the 4×4 puzzle, with 3 or 4 pairs of colored endpoints given.
Your task is to determine if a given puzzle is solvable or not.
輸入
The input consists of 4 lines, each line containing 4 characters. Each character is from the set {R, G,B, Y,W}whereW denotes the blank cells and the other characters denote endpoints with the specified color. You are guaranteed that there will be exactly 3 or 4 pairs of colored cells. If there are 3 colors in the grid, Y will be omitted.
輸出
On a single line output either “solvable” or “not solvable” (without the quotes).
樣例輸入
RGBW
WWWW
RGBY
YWWW
樣例輸出
solvable

開始讀錯了題，沒注意要連上所有的塊。
暴力枚舉每個W塊的顏色,每種情況下從起點廣搜終點，不用vis數組剪枝而採用記錄路徑（狀壓）可以保證搜到每種路徑，搜到終點時判斷是否走過了所有相同顏色的格子。
第二天仔細想想，還是寫的太麻煩了，實際上直接深搜就好了…

#define IN_LB() freopen("F:\\in.txt","r",stdin)
#define IN_PC() freopen("C:\\Users\\hz\\Desktop\\in.txt","r",stdin)
#include <bits/stdc++.h>

using namespace std;
const int dirx[] = {-1,0,1,0};
const int diry[] = {0,1,0,-1};
const char col[] ="RGBY";
char mapp[5][5];

struct node {
    int x,y,route,step;
    node() {}
    node(int x,int y,int route,int step):x(x),y(y),route(route),step(step) {}
} cR[2],cG[2],cB[2],cY[2];

int cntR,cntG,cntB,cntY;
vector <int> v;

int xytoInd(int x,int y){
    return x*4+y;
}

bool bfs(char color,node st,node ed,int cnum) {
    queue<node> q;
    q.push(st);
    while(!q.empty()) {
        int x = q.front().x,y=q.front().y,route = q.front().route,step = q.front().step;
        q.pop();
        if(x==ed.x&&y==ed.y&&cnum+1==step)return true;
        for(int i=0; i<4; i++) {
            int xx = x+dirx[i],yy=y+diry[i];
            if(xx>=0&&xx<4&&yy>=0&&yy<4&&mapp[xx][yy]==color&&((route&(1<<xytoInd(xx,yy)))==0)) {
                q.push(node(xx,yy,route|(1<<xytoInd(xx,yy)),step+1));
            }
        }
    }
    return false;
}

int main() {
//    IN_PC();
    for(int i=0; i<4; i++) {
        scanf("%s",mapp[i]);
    }
    for(int i=0; i<4; i++) {
        for(int j=0; j<4; j++) {
            if(mapp[i][j]=='W') {
                v.push_back(i*4+j);
            }
            if(mapp[i][j]=='R') {
                cR[cntR++] = node(i,j,1<<xytoInd(i,j),0);
            }
            if(mapp[i][j]=='G') {
                cG[cntG++] = node(i,j,1<<xytoInd(i,j),0);
            }
            if(mapp[i][j]=='B') {
                cB[cntB++] = node(i,j,1<<xytoInd(i,j),0);
            }
            if(mapp[i][j]=='Y') {
                cY[cntY++] = node(i,j,1<<xytoInd(i,j),0);
            }
        }
    }
    int num = v.size();
    int flag = 0;
    if(num==8) {
        int sumsta = pow(4,8);
        for(int i=0; i<sumsta; i++) {
            int c = i,cn[4]={0};
            for(int j=0; j<8; j++) {
                int x = v[j]/4,y = v[j]%4;
                mapp[x][y] = col[c%4];
                cn[c%4]++;
                c/=4;
            }
            if(bfs('R',cR[0],cR[1],cn[0])
                    &&bfs('G',cG[0],cG[1],cn[1])
                    &&bfs('B',cB[0],cB[1],cn[2])
                    &&bfs('Y',cY[0],cY[1],cn[3])) {
                flag = 1;
                break;
            }
        }
    } else if(num==10) {
        int sumsta = pow(3,10);
        for(int i=0; i<sumsta; i++) {
            int c = i,cn[3]={0};
            for(int j=0; j<10; j++) {
                int x = v[j]/4,y = v[j]%4;
                mapp[x][y] = col[c%3];
                cn[c%3]++;
                c/=3;
            }
            if(bfs('R',cR[0],cR[1],cn[0])
                    &&bfs('G',cG[0],cG[1],cn[1])
                    &&bfs('B',cB[0],cB[1],cn[2])) {
                flag = 1;
                break;
            }
        }
    }
    if(flag)printf("solvable\n");
    else printf("not solvable\n");
    return 0;
}