Time Limit:250MS Memory Limit:4096KB 64bit IO Format:%I64d & %I64u
Description
Dominoes – game played with small, rectangular blocks of wood or other material, each identified by a number of dots, or pips, on its face. The blocks usually are called bones, dominoes, or pieces and sometimes men, stones, or even cards.
The face of each piece is divided, by a line or ridge, into two squares, each of which is marked as would be a pair of dice...
The principle in nearly all modern dominoes games is to match one end of a piece to another that is identically or reciprocally numbered.
ENCYCLOPÆDIA BRITANNICA
Given a set of domino pieces where each side is marked with two digits from 0 to 6. Your task is to arrange pieces in a line such way, that they touch through equal marked sides. It is possible to rotate pieces changing left and right side.
Input
The first line of the input contains a single integer N (1 ≤ N ≤ 100) representing the total number of pieces in the domino set. The following N lines describe pieces. Each piece is represented on a separate line in a form of two digits from 0 to 6 separated by a space.
Output
Write “No solution” if it is impossible to arrange them described way. If it is possible, write any of way. Pieces must be written in left-to-right order. Every of N lines must contains number of current domino piece and sign “+” or “-“ (first means that you not rotate that piece, and second if you rotate it).
Sample Input
5
1 2
2 4
2 4
6 4
2 1
Sample Output
2 -
5 +
1 +
3 +
4 - //这个题目做了我一个晚上，有点头晕，还是要让自己保持冷静啊。。。 //多米诺骨牌有两个面，每个面有一个数字，相同的数字只能推倒相同数字面的骨牌，骨牌可以转向，给出的数字是严格的正面反面，问是否能推到所有的骨牌。 //如果可以输出骨牌的顺序和是否需要反转骨牌。如果不可以全部推倒，输出"No solution" //AC代码：
#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"string"
#include"cstdio"
#include"vector"
#include"cmath"
#include"queue"
using namespace std;
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define MX 401 bool vis[MX];
int Head[MX],cnt;
int indegree[MX],p[MX],siz,bin;
int fid[7]; struct Edge {
int v,nxt;
int flag;
} E[MX]; int find(int x) {
return fid[x]==x?x:(fid[x]=find(fid[x]));
} void union_root(int st,int ed) {
int rt1=find(st);
int rt2=find(ed);
if(rt1!=rt2)fid[rt2]=rt1;
} void edge_init() {
cnt=0;
siz=0;
memset(p,0);
memset(E,0);
memset(vis,0);
memset(Head,-1);
memset(indegree,0);
for(int i=0; i<=6; i++) {
fid[i]=i;
}
} void edge_add(int st,int ed,int flag) {
E[cnt].v=ed;
E[cnt].flag=flag;
E[cnt].nxt=Head[st];
Head[st]=cnt++;
} bool check() {
bin=-1;
int tot=0;
for(int i=0; i<=6; i++) {
if(vis[i]) {
if(bin==-1)bin=i;
if(indegree[i]%2) {
tot++;
if(tot>2)return 0;//判断奇数有多少个，超过两个奇数就判断不是
bin=i;
}
}
}
for(int i=0; i<=6; i++) {
if(bin!=-1)if(indegree[i]&&find(i)!=find(bin))return 0; //如果不是联通块就判断不是
}
if(bin==-1)return 0;
if(tot==1)return 0; //如果奇数只有一个判断不是
else return 1;
} void Fleury(int u,int e) {
for(int i=Head[u]; ~i; i=E[i].nxt) {
int v=E[i].v;
if(vis[i|1])continue;
vis[i|1]=1;
Fleury(v,E[i].flag);
}
if(e)p[siz++]=e; //标记是哪一条边。
} int main() {
int m,st,ed;
while(~scanf("%d",&m)) {
edge_init();
for(int i=1; i<=m; i++) {
scanf("%d%d",&st,&ed);
vis[st]=vis[ed]=1;
edge_add(st,ed,i); //正向
edge_add(ed,st,-i); //记录多米诺骨牌是否需要转向 反向
indegree[st]++;
indegree[ed]++;
union_root(st,ed); //合成联通块
}
bool flag=check();
if(!flag)puts("No solution");
else {
memset(vis,0); //再次清空标记数组
Fleury(bin,0);
for(int i=siz-1; i>=0; i--) {
printf("%d %c\n",p[i]>0?p[i]:-p[i],p[i]>0?'+':'-');
}
}
}
return 0;
}