1251 - Forming the Council
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| Time Limit: 2 second(s) | Memory Limit: 32 MB |
In a city there are n voters, and m people formed the Govt. council. The council members are numbered from 1 to m. Now everyone is complaining that the council is biased. So, they made a
plan. The plan is that the voters are given a chance to vote again to form the new council. A vote will be like ±i ±j. ‘+‘ means the voter wants that member to be in the council, ‘-‘ means the voter doesn‘t
want the member to be in the council. For example, there are 4 voters, they voted like
+1 -3 the voter wants member 1 to be kept in the council or member 3 to be thrown out
+2 +3 the voter wants member 2 to be kept in the council or member 3 to be kept in the council
-1 -2 the voter wants member 1 to be thrown out or member 2 to be thrown out
-4 +1 the voter wants member 4 to be thrown out or member 1 to be kept in the council
A voter will be satisfied if at least one of his wishes becomes true. Now your task is to form the council such that all the voters are happy.
Input
Input starts with an integer T (≤ 20), denoting the number of test cases.
Each case starts with a line containing two integers n (1 ≤ n ≤ 20000) and m (1 ≤ m ≤ 8000). Each of the next n lines contains a vote in the form ±i ±j (1 ≤ i, j ≤ m).
Output
For each case, print the case number and ‘Yes‘ if a solution exists, or ‘No‘ if there is no solution. Then if the result is yes, print another line containing the number of members in the council followed by the members
in ascending order. And print a single space between two numbers. There can be many solutions. Any valid one will do.
Sample Input |
Output for Sample Input |
|
3 4 3 +1 +3 +2 -1 +2 -3 -1 -2 4 2 +1 -2 +1 +2 -1 -2 -1 +2 1 3 +1 -3 |
Case 1: Yes 2 2 3 Case 2: No Case 3: Yes 0 |
Note
This is a special judge problem. Wrong output format may cause wrong answer.
PROBLEM SETTER: JANE ALAM JAN
题意：一个城市的理事会有M个投票人和N个公民组成，当中公民编号从1——N。
如今要建立一个新的理事会，
每一个投票人都给出了自己的意见，如
1，+i +j 表示 i公民 和 j公民 至少有一个留在理事会
2，+i -j 表示 i公民留在理事会 和 j公民离开理事会 至少有一个成立
3。-i +j 表示 i公民离开理事会 和 j公民留在理事会 至少有一个成立
4，-i -j 表示 i公民 和 j公民 至少有一个离开理事会
如今让你找出一种方案选取若干个公民留在理事会。
若不存在方案输出No。
存在输出Yes，在下一行输出选择的公民总数 并输出被选择公民的编号。
思路：2-sat 推断可行解 + 反向拓扑染色输出可行解。不是非常难的题目，没什么好说的。这里仅仅说下建图。
建图：用 i 表示 i 留在理事会，i + N表示 i 离开理事会。
1。+i +j 表示 i公民 和 j公民 至少有一个留在理事会
addEdge(j + N, i);// j 离开 那么 i 必然留下
addEdge(i + N, j);// i 离开 那么 j 必然留下
2。+i -j 表示 i公民留在理事会 和 j公民离开理事会 至少有一个成立
addEdge(j, i);// j 留下 i 必然留下
addEdge(i + N, j + N);// i 离开 j 必然离开
3，-i +j 表示 i公民离开理事会 和 j公民留在理事会 至少有一个成立
addEdge(j + N, i + N);// j 离开 i 必然离开
addEdge(i, j);// i 留下 j 必然留下
4。-i -j 表示 i公民 和 j公民 至少有一个离开理事会
addEdge(j, i + N);// j 留下 那么 i 必然离开
addEdge(i, j + N);// i 留下 那么 j 必然离开
AC代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#define MAXN 16000+10
#define MAXM 40000+10
using namespace std;
struct Edge
{
int from, to, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int low[MAXN], dfn[MAXN];
int sccno[MAXN], scc_cnt;
int dfs_clock;
stack<int> S;
bool Instack[MAXN];
int N, M;
void init()
{
edgenum = 0;
memset(head, -1, sizeof(head));
}
void addEdge(int u, int v)
{
Edge E = {u, v, head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
void getMap()
{
int a, b;
while(M--)
{
scanf("%d%d", &a, &b);
if(a > 0 && b > 0)//a 和 b 至少一个留下
{
addEdge(b + N, a);
addEdge(a + N, b);
}
else if(a > 0 && b < 0)//a留 和 b走 至少成立一个
{
b = -b;
addEdge(b, a);
addEdge(a + N, b + N);
}
else if(a < 0 && b > 0)//a走 和 b留 至少成立一个
{
a = -a;
addEdge(b + N, a + N);
addEdge(a, b);
}
else//a 和 b 至少走一个
{
a = -a, b = -b;
addEdge(b, a + N);
addEdge(a, b + N);
}
}
}
void tarjan(int u, int fa)
{
int v;
low[u] = dfn[u] = ++dfs_clock;
S.push(u);
Instack[u] = true;
for(int i = head[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if(!dfn[v])
{
tarjan(v, u);
low[u] = min(low[u], low[v]);
}
else if(Instack[v])
low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u])
{
scc_cnt++;
for(;;)
{
v = S.top();S.pop();
Instack[v] = false;
sccno[v] = scc_cnt;
if(v == u) break;
}
}
}
void find_cut(int l, int r)
{
memset(low, 0, sizeof(low));
memset(dfn, 0, sizeof(dfn));
memset(sccno, 0, sizeof(sccno));
memset(Instack, false, sizeof(Instack));
dfs_clock = scc_cnt = 0;
for(int i = l; i <= r; i++)
if(!dfn[i]) tarjan(i, -1);
}
vector<int> G[MAXN];
int in[MAXN];
void suodian()//反向建图
{
for(int i = 1; i <= scc_cnt; i++) G[i].clear(), in[i] = 0;
for(int i = 0; i < edgenum; i++)
{
int u = sccno[edge[i].from];
int v = sccno[edge[i].to];
if(u != v)
G[v].push_back(u), in[u]++;
}
}
int k = 1;
int fp[MAXN];//建立SCC到SCC的映射
int color[MAXN];//染色
void toposort()
{
memset(color, -1, sizeof(color));
queue<int> Q;
for(int i = 1; i <= scc_cnt; i++) if(in[i] == 0) Q.push(i);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
if(color[u] == -1)
{
color[u] = 1;
color[fp[u]] = 0;
}
for(int i = 0; i < G[u].size(); i++)
{
int v = G[u][i];
if(--in[v] == 0)
Q.push(v);
}
}
}
void solve()
{
printf("Case %d: ", k++);
for(int i = 1; i <= N; i++)
{
if(sccno[i] == sccno[i+N])
{
printf("No\n");
return ;
}
else
{
fp[sccno[i]] = sccno[i+N];
fp[sccno[i+N]] = sccno[i];
}
}
printf("Yes\n");
suodian();
toposort();//反向拓扑
int ans = 0;
for(int i = 1; i <= N; i++)
{
if(color[sccno[i]] == 1)
ans++;
}
printf("%d", ans);
for(int i = 1; i <= N; i++)
{
if(color[sccno[i]] == 1)
printf(" %d", i);
}
printf("\n");
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &M, &N);
init();
getMap();
find_cut(1, 2*N);
solve();
}
return 0;
}