标签：ems case item mina pop ever div class must
InputThe 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 B
1,B
2, ...,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.OutputFor 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
1 #include<stack> 2 #include<cstdio> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #define mem(array) memset(array,0,sizeof(array)) 7 using namespace std; 8 9 const int maxn = 60005; 10 const int maxm = 999999; 11 12 stack<int> S; 13 14 struct node { 15 int to, next; 16 }e[maxm]; 17 18 int k, n, m, tot, index; 19 int Low[maxn], Dfn[maxn], head[maxn], cmp[maxn]; 20 bool use[maxn]; 21 22 void Inite() { 23 k = tot = index = 0; 24 mem(Dfn); mem(use); mem(cmp); 25 memset(head, -1, sizeof(head)); 26 } 27 28 void addedge(int u, int v) { 29 e[tot].to = v; 30 e[tot].next = head[u]; 31 head[u] = tot++; 32 } 33 34 void Tarjan(int u) { 35 Low[u] = Dfn[u] = ++index; 36 S.push(u); 37 use[u] = 1; 38 for (int i = head[u]; i != -1; i = e[i].next) { 39 int v = e[i].to; 40 if (!Dfn[v]) { 41 Tarjan(v); 42 Low[u] = min(Low[u], Low[v]); 43 } 44 else if (use[v]) { 45 Low[u] = min(Low[u], Dfn[v]); 46 } 47 } 48 int v; 49 if (Dfn[u] == Low[u]) { 50 k++; 51 do { 52 v = S.top(); 53 S.pop(); 54 cmp[v] = k; 55 use[v] = 0; 56 } while (v != u); 57 } 58 } 59 60 int main() 61 { 62 int kase; 63 scanf("%d", &kase); 64 for (int t = 1; t <= kase; t++) { 65 scanf("%d%d", &n, &m); 66 Inite(); 67 for (int i = 0; i < n; i++) { 68 addedge(i * 6 + 0, i * 6 + 3); 69 addedge(i * 6 + 0, i * 6 + 5); 70 addedge(i * 6 + 2, i * 6 + 1); 71 addedge(i * 6 + 2, i * 6 + 5); 72 addedge(i * 6 + 4, i * 6 + 1); 73 addedge(i * 6 + 4, i * 6 + 3); 74 } 75 for (int i = 0; i < n; i++) { 76 int v; 77 scanf("%d", &v); 78 if (v == 1) { 79 addedge(i * 6 + 3, i * 6 + 0); 80 addedge(i * 6 + 1, i * 6 + 2); 81 } 82 else if (v == 2) { 83 addedge(i * 6 + 5, i * 6 + 2); 84 addedge(i * 6 + 3, i * 6 + 4); 85 } 86 else { 87 addedge(i * 6 + 5, i * 6 + 0); 88 addedge(i * 6 + 1, i * 6 + 4); 89 } 90 } 91 for (int i = 0; i < m; i++) { 92 int a, b, c; 93 scanf("%d%d%d", &a, &b, &c); 94 a--; b--; 95 if (c) { 96 addedge(a * 6 + 0, b * 6 + 1); 97 addedge(a * 6 + 2, b * 6 + 3); 98 addedge(a * 6 + 4, b * 6 + 5); 99 addedge(b * 6 + 0, a * 6 + 1); 100 addedge(b * 6 + 2, a * 6 + 3); 101 addedge(b * 6 + 4, a * 6 + 5); 102 } 103 else { 104 addedge(a * 6 + 0, b * 6 + 0); 105 addedge(a * 6 + 2, b * 6 + 2); 106 addedge(a * 6 + 4, b * 6 + 4); 107 addedge(b * 6 + 0, a * 6 + 0); 108 addedge(b * 6 + 2, a * 6 + 2); 109 addedge(b * 6 + 4, a * 6 + 4); 110 } 111 } 112 bool flag = false; 113 printf("Case #%d: ", t); 114 for (int i = 0; i < 6 * n; i++) if (!Dfn[i]) Tarjan(i); 115 for (int i = 0; i < 3 * n; i++) if (cmp[i << 1] == cmp[(i << 1) | 1]) flag = true; 116 if (flag) printf("no\n"); 117 else printf("yes\n"); 118 } 119 return 0; 120 }
Eliminate the Conflict HDU - 4115
标签：ems case item mina pop ever div class must
原文地址：http://www.cnblogs.com/zgglj-com/p/8000136.html
