开始读错了题，没注意要连上所有的块。
暴力枚举每个W块的颜色,每种情况下从起点广搜终点，不用vis数组剪枝而采用记录路径（状压）可以保证搜到每种路径，搜到终点时判断是否走过了所有相同颜色的格子。
第二天仔细想想，还是写的太麻烦了，实际上直接深搜就好了…
#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;
}