Eliminate the Conflict（2 - sat）

Eliminate the Conflict

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1838    Accepted Submission(s): 789

Problem Description

Conflicts are everywhere in the world, from the young to the elderly, from families to countries. Conflicts cause quarrels, fights or even wars. How wonderful the world will be if all conflicts can be eliminated.
Edward contributes his lifetime to invent a 'Conflict Resolution Terminal' and he has finally succeeded. This magic item has the ability to eliminate all the conflicts. It works like this:
If any two people have conflict, they should simply put their hands into the 'Conflict Resolution Terminal' (which is simply a plastic tube). Then they play 'Rock, Paper and Scissors' in it. After they have decided what they will play, the tube should be opened and no one will have the chance to change. Finally, the winner have the right to rule and the loser should obey it. Conflict Eliminated!
But the game is not that fair, because people may be following some patterns when they play, and if the pattern is founded by others, the others will win definitely.
Alice and Bob always have conflicts with each other so they use the 'Conflict Resolution Terminal' a lot. Sadly for Bob, Alice found his pattern and can predict how Bob plays precisely. She is very kind that doesn't want to take advantage of that. So she tells Bob about it and they come up with a new way of eliminate the conflict:
They will play the 'Rock, Paper and Scissors' for N round. Bob will set up some restricts on Alice.
But the restrict can only be in the form of "you must play the same (or different) on the ith and jth rounds". If Alice loses in any round or break any of the rules she loses, otherwise she wins.
Will Alice have a chance to win?

 

 

Input

The first line contains an integer T(1 <= T <= 50), indicating the number of test cases.
Each test case contains several lines.
The first line contains two integers N,M(1 <= N <= 10000, 1 <= M <= 10000), representing how many round they will play and how many restricts are there for Alice.
The next line contains N integers B₁,B₂, ...,B_N, where B_i represents what item Bob will play in the i^th round. 1 represents Rock, 2 represents Paper, 3 represents Scissors.
The following M lines each contains three integers A,B,K(1 <= A,B <= N,K = 0 or 1) represent a restrict for Alice. If K equals 0, Alice must play the same on A^th and B^th round. If K equals 1, she must play different items on Ath and Bthround.

 

 

Output

For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y is "yes" or "no" represents whether Alice has a chance to win.

 

 

Sample Input

2

3 3

1 1 1

1 2 1

1 3 1

2 3 1

5 5

1 2 3 2 1

1 2 1

1 3 1

1 4 1

1 5 1

2 3 0

 

 

Sample Output

Case #1: no

Case #2: yes

Hint

'Rock, Paper and Scissors' is a game which played by two person. They should play Rock, Paper or Scissors by their hands at the same time. Rock defeats scissors, scissors defeats paper and paper defeats rock. If two people play the same item, the game is tied..

 

 

Source

 

2011 Asia ChengDu Regional Contest 

 

     题意：

     给出 T 组样例，后给出 n，m，代表有 n round 剪刀石头布游戏，除了不输掉任意 1 round之外，还要满足 m 个条件。每个条件给出 a，b，k。如果 k == 0，代表要求 a 和 b 要一样，k == 1 的话则要求 a 和 b 不一样。输出能不能赢。

 

     思路：

     2 - sat。预处理出每一局能出的两个数是，令其一个为真一个为假。当 k == 1 的时候，两个数相等的时候则要求建边。当 k == 0 的时候，两个数不相等的时候需要建边。最后走一遍 2 - sat 则可得出答案。

 

     AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>

using namespace std;

const int MAX = 10005 * 2;

int n, m;
int res[MAX][2];

vector<int> v[MAX];

int dfs_clock, scc_cnt;
int pre[MAX], low[MAX], cmp[MAX];
stack<int> s;

void init () {
    for (int i = 0; i < MAX; ++i) v[i].clear();
}

void dfs(int u) {
    pre[u] = low[u] = ++dfs_clock;
    s.push(u);

    for (int i = 0; i < v[u].size(); ++i) {
        if (!pre[ v[u][i] ]) {
            dfs( v[u][i] );
            low[u] = min(low[u], low[ v[u][i] ]);
        } else if (!cmp[ v[u][i] ]) {
            low[u] = min(low[u], pre[ v[u][i] ]);
        }
    }

    if (low[u] == pre[u]) {
        ++scc_cnt;
        for (;;) {
            int x = s.top(); s.pop();
            cmp[x] = scc_cnt;
            if (x == u) break;
        }
    }
}

void scc() {
    dfs_clock = scc_cnt = 0;
    memset(pre, 0, sizeof(pre));
    memset(cmp, 0, sizeof(cmp));

    for (int u = 1; u <= 2 * n; ++u) {
        if (!pre[u]) dfs(u);
    }
}

int main() {

    int t;
    scanf("%d", &t);

    for (int tt = 1; tt <= t; ++tt) {
        init();

        scanf("%d%d", &n, &m);

        for (int i = 1; i <= n; ++i) {
            int ans;
            scanf("%d", &ans);
            if (ans == 1) {
                res[i][0] = 1, res[i][1] = 2;
            } else if (ans == 2) {
                res[i][0] = 2, res[i][1] = 3;
            } else {
                res[i][0] = 3, res[i][1] = 1;
            }
        }

        while (m--) {
            int a, b, k;
            scanf("%d%d%d", &a, &b, &k);
            if (k) {
                if (res[a][0] == res[b][0]) {
                    v[a].push_back(b + n);
                    v[b].push_back(a + n);
                }
                if (res[a][0] == res[b][1]) {
                    v[a].push_back(b);
                    v[b + n].push_back(a + n);
                }
                if (res[a][1] == res[b][0]) {
                    v[a + n].push_back(b + n);
                    v[b].push_back(a);
                }
                if (res[a][1] == res[b][1]) {
                    v[a + n].push_back(b);
                    v[b + n].push_back(a);
                }
            } else {
                if (res[a][0] != res[b][0]) {
                    v[a].push_back(b + n);
                    v[b].push_back(a + n);
                }
                if (res[a][0] != res[b][1]) {
                    v[a].push_back(b);
                    v[b + n].push_back(a + n);
                }
                if (res[a][1] != res[b][0]) {
                    v[a + n].push_back(b + n);
                    v[b].push_back(a);
                }
                if (res[a][1] != res[b][1]) {
                    v[a + n].push_back(b);
                    v[b + n].push_back(a);
                }
            }
        }

        scc();

        printf("Case #%d: ", tt);

        int temp = 0;
        for (int i = 1; i <= n; ++i) {
            if (cmp[i] == cmp[n + i]) {
                printf("no\n");
                temp = 1;
                break;
            }
        }

        if (!temp) printf("yes\n");

    }


    return 0;
}