Light 1003 - Drunk (拓扑排序)

题意

如果要喝一种饮料，需要先喝另一种饮料。
给出这种关系，问一个人能不能喝完全部的饮料。

思路

一开始我以为只能选一种饮料喝，然后喝完全部的饮料。

WA了几次后才发现题意就是能不能喝完，也就是说会不会出现不能被喝的饮料，也就是成环的饮料。

用拓扑排序去搞。

代码

#include <stack>#include <cstdio>#include <list>#include <cassert>#include <set>#include <fstream>#include <iostream>#include <string>#include <vector>#include <queue>#include <functional>#include <cstring>#include <algorithm>#include <cctype>//#pragma comment(linker, "/STACK:102400000,102400000")#include <string>#include <map>#include <cmath>#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/hash_policy.hpp>using namespace std;using namespace __gnu_pbds;#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define X first#define Y second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define LC rt << 1, l, mid#define RC rt << 1|1, mid + 1, r#define LRT rt << 1#define RRT rt << 1|1#define FOR(i, a, b) for (int i=(a); (i) < (b); (i)++)#define FOOR(i, a, b) for (int i = (a); (i)<=(b); (i)++)const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const double eps = 1e-8;const int MAXN = 2e4+10;const int MOD = 1e9;const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };const int seed = 131;int cases = 0;typedef pair<int, int> pii;map<string, int> mp;int clk, in[MAXN];vector<int> G[MAXN];bool Toposort(){    queue<int> Q;    int cnt = 0;    FOOR(i, 1, clk)        if (in[i] == 0)            Q.push(i);    while (!Q.empty())    {        int u = Q.front(); Q.pop();        FOR(i, 0, SZ(G[u]))        {            int v = G[u][i];            in[v]--;            if (in[v] == 0)                Q.push(v);        }        cnt++;    }    return cnt == clk;}int main(){    //ROP;    int T;    scanf("%d", &T);    while (T--)    {        MS(in, 0);        mp.clear(); clk = 0;        int n;        scanf("%d", &n);        FOOR(i, 0, 2*n) G[i].clear();        while (n--)        {            string a, b;            cin >> a >> b;            if (!mp.count(a)) mp[a] = ++clk;            if (!mp.count(b)) mp[b] = ++clk;            G[mp[b]].PB(mp[a]);            in[mp[a]]++;        }        printf("Case %d: %s\n", ++cases, Toposort() ? "Yes" : "No");    }    return 0;}