HDU 1285 确定比赛名次 拓扑排序

思路：就是每次找一个入度为0的点。然后把这个点删掉，并将这个点连着的其他点的入度减1。

http://acm.hdu.edu.cn/showproblem.php?pid=1285

/*********************************************    Problem : HDU 1258    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 505;const int MAXE = 50005;typedef long long LL;typedef unsigned long long ULL;int T,n,m,k;int fx[] = {0,1,-1,0,0};int fy[] = {0,0,0,-1,1};int ru[505];struct Edge { //记录边    int to;    Edge * next;}E[MAXE],*EE;struct Gragh { //记录图的结点    Edge * first;}G[MAXN];void addedge(int u,int v) { //加边,双向边    EE->to = v ; EE -> next = G[u].first ; G[u].first = EE ++;    //EE->to = u ; EE -> next = G[v].first ; G[v].first = EE ++;}int ans[505];void init() {    EE = E;    cls(G,0); cls(ru,0);}void input() {    int u,v;    rep(i,1,m) {        scanf("%d %d",&u,&v);        addedge(u,v);        ru[v] ++;    }}void JIAN(int u) {    repE(p,u) {        int v = p -> to;        ru[v] --;    }}void solve() {    rep(i,1,n)    rep(j,1,n) {        if(ru[j] == 0) {            ru[j] = -1;            ans[i] = j;            JIAN(j);            break;        }    }    rep(i,1,n) {        printf("%d",ans[i]);        if(i == n) printf("\n");        else printf(" ");    }}int main(void) {    //freopen("a.in","r",stdin);    //scanf("%d",&T); while(T--) {    while(~scanf("%d %d",&n,&m)) {    //while(~scanf("%d",&n)) {        init();        input();        solve();    }    return 0;}