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  • [JSOI2010]满汉全席 -- 2-SAT

    2-SAT快忘了,回忆了一下 x--->y   代表选择x必选择y

    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<stack>
    #include<cstdio>
    using namespace std;
    const int maxn = 100000;
    struct Node {
    	int to;
    	int next;
    }G[maxn];
    
    int head[maxn];
    int vis[maxn];
    int z;
    void add(int be, int en) {
    	G[++z].to = en;
    	G[z].next = head[be];
    	head[be] = z;
    }
    int n, m;
    
    int get(char *cn) {
    	int ret = 0;
    	for (int i = 1; cn[i]; i++) {
    		if (cn[i] >= '0' && cn[i] <= '9') {
    			ret *= 10;
    			ret += cn[i] - '0';
    		}
    		else break;
    	}
    	return ret;
    }
    int dfn[maxn], clor[maxn], low[maxn];
    int df, ans;
    stack<int>s;
    
    
    int tarjan(int x) {
    	dfn[x] = low[x] = ++df;
    	s.push(x);
    	vis[x] = 1;
    	for (int i = head[x]; i; i = G[i].next) {
    		int p = G[i].to;
    		if (!dfn[p]) {
    			tarjan(p);
    			low[x] = min(low[x], low[p]);
    		}
    		else if (vis[p]) {
    			low[x] = min(low[x], dfn[p]);
    		}
    	}
    	if (low[x] == dfn[x]) {
    		ans++;
    		while (1) {
    			int c = s.top();
    			s.pop();
    			vis[c] = 0;
    			clor[c] = ans;
    			if (c == x) break;
    		}
    	}
    	return 0;
    }
    int main() {
    	int t;
    	scanf("%d", &t);
    	while (t--) {
    		
    		char a[100];
    		char b[100];
    		memset(head, 0, sizeof(head));
    		memset(clor, 0, sizeof(clor));
    		memset(dfn, 0, sizeof(dfn));
    		memset(low, 0, sizeof(low));
    		df = 0;
    		ans = 0;
    		z = 0;
    		scanf("%d%d", &n, &m);
    		for (int i = 0; i < m; i++) {//m+n,h正
    			scanf("%s%s", a, b);
    			int be = get(a);
    			int en = get(b);
    
    			if (a[0] == 'm' && b[0] == 'm') {
    				add(be, en + n);//h--m
    				add(en, be + n);//
    			}
    			else if (a[0] == 'h' && b[0] == 'h') {
    				add(be + n, en);
    				add(en + n, be);
    			}
    			else if (a[0] == 'm' && b[0] == 'h') {
    				add(be, en);
    				add(en + n, be + n);
    			}
    			else if (a[0] == 'h' && b[0] == 'm') {
    				add(be + n, en + n);
    				add(en, be);
    			}
    
    
    			a[0] = 0;
    			b[0] = 0;
    		}
    		for (int i = 1; i <= 2*n; i++) {
    			if (!dfn[i]) tarjan(i);
    		}
    
    		int flag = 0;
    		for (int i = 1; i <= n ; i++) {
    			if (clor[i] == clor[i + n]) {
    				flag = 1;
    				break;
    			}
    		}
    		if (!flag) printf("GOOD
    ");
    		else printf("BAD
    ");
    		while (s.size()) s.pop();
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/lesning/p/12526450.html
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