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【拓扑排序】 HDOJ 3342 Legal or Not

2014-08-17 11:01 309 查看
水拓扑。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 105
#define maxm 10005
#define eps 1e-10
#define mod 998244353
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
//typedef int LL;
using namespace std;
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
// head

int h[maxn], next[maxm], v[maxm];
int in[maxn], cnt, n, m, ok;
queue<int> q;

void init(void)
{
cnt = 0;
memset(h, -1, sizeof h);
memset(in, 0, sizeof in);
}
void addedges(int u, int vv)
{
next[cnt] = h[u], h[u] = cnt, v[cnt] = vv, cnt++;
}
void read(void)
{
int a, b;
ok = 1;
while(m--) {
//scanf("%d%d", &a, &b);
scanf(a), scanf(b);
if(a == b) ok = 0;
addedges(a, b), in[b]++;
}
}
void work(void)
{
m = 0;
for(int i = 0; i < n; i++) if(!in[i]) q.push(i), m++;
while(!q.empty()) {
int u = q.front(); q.pop();
for(int e = h[u]; ~e; e = next[e]) {
in[v[e]]--;
if(!in[v[e]]) q.push(v[e]), m++;
}
}
if(m == n) printf("YES\n");
else printf("NO\n");
}
int main(void)
{
while(scanf("%d%d", &n, &m), n != 0 || m != 0) {
init();
read();
if(ok) work();
else printf("NO\n");
}
return 0;
}
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