UVA 10731Test （scc）

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题意:

求最少的集合个数,使得集合内元素都是冲突的,有序输出

分析:

都冲突就是都相互可达嘛,有向图就是scc咯,然后用string排序输出

代码:

////  Created by TaoSama on 2015-11-08//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 26, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;vector<int> G[N];int dfn[N], low[N], in[N], stk[N], top, dfsNum, scc;string ans[26];void tarjan(int u) {    low[u] = dfn[u] = ++dfsNum;    in[u] = true;    stk[++top] = u;    for(auto v : G[u]) {        if(!dfn[v]) {            tarjan(v);            low[u] = min(low[u], low[v]);        } else if(in[v]) low[u] = min(low[u], dfn[v]);    }    if(low[u] == dfn[u]) {        ans[++scc] = "";        while(true) {            int v = stk[top--];            in[v] = false;            ans[scc] += char(v + 'A');            if(v == u) break;        }        sort(ans[scc].begin(), ans[scc].end());    }}bool vis[26];void init() {    top = scc = dfsNum = 0;    memset(dfn, 0, sizeof dfn);    memset(low, 0, sizeof low);    memset(in, 0, sizeof in);    memset(vis, 0, sizeof vis);    for(int i = 0; i < 26; ++i) G[i].clear();}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int kase = 0;    while(scanf("%d", &n) == 1 && n) {        init();        char buf[10][2];        for(int i = 1; i <= n; ++i) {            for(int j = 1; j <= 6; ++j) scanf("%s", buf[j]);            int u = buf[6][0] - 'A', v;            vis[u] = true;            for(int j = 1; j <= 5; ++j) {                v = buf[j][0] - 'A';                vis[v] = true;                if(v == u) continue;                G[u].push_back(v);            }        }        for(int i = 0; i < 26; ++i) if(vis[i] && !dfn[i]) tarjan(i);        sort(ans + 1, ans + 1 + scc);        if(kase++) puts("");        for(int i = 1; i <= scc; ++i)            for(int j = 0; j < ans[i].size(); ++j)                printf("%c%c", ans[i][j], " \n"[j == ans[i].size() - 1]);    }    return 0;}

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