【每日一题】 UVA

一开始用set存xjb分类讨论，然后wa，

然后简化了一点，改用vector，然wa

最后又发现没有初始化，然wa

wa了一个半小时

最后看了题解orz

然后找了一组样例把自己的代码改对了

/*

1 1
1 1
1 1
1 1
1 2
1 2

*/

正统题解：不妨设三条边为a<=b<=c， 那么对每个面（边对）sort后，必然得到 ab ab ac ac bc bc 然后照着这个序列写六个判断就好了orz

#define _CRT_SECURE_NO_WARNINGS
#include<cmath>
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<stack>
#include<vector>
#include<string.h>
#include<queue>
#include<string>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
#define mod 1000000007
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
#define eps 1e-6
#define pb push_back

#define mp make_pair
#define x first
#define y second
pair<int, int> a[7];
int main() {
    while (cin >> a[1].x >> a[1].y){
        
        rep(i, 2, 6)scanf("%d%d", &a[i].x, &a[i].y);
        rep(i, 1, 6)if(a[i].x>a[i].y)swap(a[i].x,a[i].y);
        sort(a + 1, a + 7);
        puts(a[1] == a[2] && a[3] == a[4] && a[5] == a[6] && a[1].x==a[3].x&&a[3].y==a[5].y&&a[5].x==a[1].y?"POSSIBLE":"IMPOSSIBLE");

}
}
/*
POSSIBLE
IMPOSSIBLE
*/

第一版代码，先判每个面出现了两遍，再把不同的三个面找出来，按有几个面是正方形分类

#define _CRT_SECURE_NO_WARNINGS
#include    <cmath>
#include <iostream>
#include    <stdio.h>
#include<algorithm>
#include        <map>
#include     <cstring>
#include      <time.h>
#include    <string>
#include    <vector>
#include <set>
using namespace std;
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
const double pi = acos(-1.0);
const int maxn = 100 + 5;
int n,m;


int main()
{

    int x, y;

    while (cin >> x >> y){
        set<int> S[7];
        vector < set<int>> V;
        S[1].insert(x), S[1].insert(y);
        rep(i, 2, 6) {
            
            cin >> x >> y;
            S[i].insert(x), S[i].insert(y);
    }
        int f[7];
        mmm(f, 0);
        int vis[7]; mmm(vis, 0);
        rep(i, 1, 6)rep(j, 1, 6)if (i != j) {
            if (S[i] == S[j]) { f[i] = 1; if (!vis[i]&&!vis[j]) { V.push_back(S[i]); vis[j] = 1; vis[i] = 1;} }
        }
        rep(i, 1, 6)if (!f[i]) { f[0] = 1; }
        if (f[0] == 1) { puts("IMPOSSIBLE"); continue; }
        int cnt=0;
        rep(i, 0, 2)if (V[i].size() == 1)cnt++;
        if (cnt == 3) { if (V[1] == V[0] && V[1] == V[2]) puts("POSSIBLE"); else puts("IMPOSSIBLE"); continue; }
        if (cnt == 1) {
            vector < set<int>> T;
            for (auto t : V)if (t.size() == 2) { T.push_back(t); }
            if(*T.begin()==*T.rbegin()) { puts("POSSIBLE"); continue; }else { puts("IMPOSSIBLE"); continue; }
        }
        if(cnt==2) { puts("IMPOSSIBLE"); continue; }
        if (cnt == 0) {
            int ok = 1;
            vector < int> T;
            rep(i, 1, 2)if (*V[0].begin() == *V[i].begin())T.push_back(*V[i].rbegin());
            else if (*V[0].begin() == *V[i].rbegin())T.push_back(*V[i].begin());
            rep(i, 1, 2)if (*V[0].rbegin() == *V[i].begin())T.push_back(*V[i].rbegin());
            else if (*V[0].rbegin() == *V[i].rbegin())T.push_back(*V[i].begin());
            //else ok = 0;
            if(T.size()!=2){ puts("IMPOSSIBLE"); continue; }
            if(T.front()!=T.back()) { puts("IMPOSSIBLE"); continue; }
            else { puts("POSSIBLE"); continue; }
        }
    }
    cin >> n;
    return 0;
}
/*

1 1
2 1
2 1
1 1
1 1
1 1
*/

第二版代码：用vector存，并且简化了分类：要么可以 要么不可能//废话

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<set>
#include<vector>
using namespace std;
typedef long long ll;
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
int main()
{
    
    int x, y;
    while (cin >> x >> y) {
        vector < vector<int>> V;
        vector<int> v[7];
        if (x > y)swap(x, y);
        v[1].push_back(x); v[1].push_back(y);
        rep(i, 2, 6) {

            cin >> x >> y;
            if (x > y)swap(x, y);
            v[i].push_back(x); v[i].push_back(y);
        }
        int f[7]; int vis[7]; mmm(vis, 0);
        mmm(f, 0);
        rep(i, 1, 6)rep(j, 1, 6)if (i != j) {
            if (v[i] == v[j]) { f[i] = 1; if (!vis[i]&&!vis[j]) { V.push_back(v[i]); vis[j] = 1; vis[i] = 1;}  }
        }
        rep(i, 1, 6)if (!f[i]) { f[0] = 1; }
        if (f[0] == 1) { puts("IMPOSSIBLE"); continue; }
        int ok = 1;
        int t = 0;
        int x = 2;
        if (V[0].front() == V[1].front())t = V[1].back(); else if (V[0].front() == V[1].back())t = V[1].front(); else {
            x = 1;
            if (V[0].front() == V[2].front())t = V[2].back(); else if (V[0].front() == V[2].back())t = V[2].front(); else ok = 0;
        }
        int t2 = 0;
        if (t == V[x].front())t2 = V[x].back(); else if (t == V[x].back())t2 = V[x].front(); else {
            ok = 0;
        }
        if (t2 != V[0].back())ok = 0;
        if (ok)puts("POSSIBLE");
        else puts("IMPOSSIBLE");
    }
    return 0;
}