HDU 5413 拓扑排序+BITSET

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5413

题意：给定一张有向图，对于边e<u,v>，如果去掉它的话，u仍然可以到达v的话，那么这条边就是多余边。

解法：显然的一个思路是倒着做。先topo一遍，然后倒着来，这里需要用到bitset维护每个点可以和哪些点直接相连，利用bitset是为了不超内存，然后用bitset处理出关系之后就可以遍历寻找答案了。

#include <bits/stdc++.h>using namespace std;struct FastIO{    static const int S = 1310720;    int wpos;    char wbuf[S];    FastIO() : wpos(0) {}    inline int xchar()    {        static char buf[S];        static int len = 0, pos = 0;        if (pos == len)            pos = 0, len = fread(buf, 1, S, stdin);        if (pos == len) return -1;        return buf[pos ++];    }    inline int xuint()    {        int c = xchar(), x = 0;        while (c <= 32) c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x;    }    inline int xint()    {        int s = 1, c = xchar(), x = 0;        while (c <= 32) c = xchar();        if (c == '-') s = -1, c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x * s;    }    inline void xstring(char *s)    {        int c = xchar();        while (c <= 32) c = xchar();        for (; c > 32; c = xchar()) * s++ = c;        *s = 0;    }    inline void wchar(int x)    {        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;        wbuf[wpos ++] = x;    }    inline void wint(int x)    {        if (x < 0) wchar('-'), x = -x;        char s[24];        int n = 0;        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;        while (n--) wchar(s[n]);        wchar('\n');    }    inline void wstring(const char *s)    {        while (*s) wchar(*s++);    }    ~FastIO()    {        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;    }} io;const int maxn = 2e4+2;int n, m, top, in[maxn], rec[maxn];bitset <maxn> dp[maxn];int head[maxn], edgecnt;struct edge{    int u, v, next;}E[100002];struct node{    int u, v;}q[100002];void add(int u, int v){    E[edgecnt].u = u ,E[edgecnt].v = v, E[edgecnt].next = head[u], head[u] = edgecnt++;}void Toposort(){    top = 0;    for(int i=1; i<=n; i++) if(in[i]==0) rec[top++]=i;    for(int i=0; i<top; i++){        int u = rec[i];        for(int j=head[u]; ~j; j=E[j].next){            int v = E[j].v;            if(--in[v]==0) rec[top++] = v;        }    }}int main(){    int T;    T = io.xint();    while(T--)    {        n = io.xint();        m = io.xint();        edgecnt=0;        for(int i=1; i<=n; i++){            head[i]=-1;            in[i]=0;        }        for(int i=1; i<=m; i++){            int u, v;            u = io.xint();            v = io.xint();            q[i].u = u, q[i].v = v;            add(u, v);            in[v]++;        }        Toposort();        for(int i=0; i<=n; i++){            dp[i].reset();            dp[i].set(i);        }        int ans = 0;        for(int i=top-1; i>=0; i--){            int u = rec[i];            for(int j=head[u]; ~j; j=E[j].next){                int v = E[j].v;                dp[u][v] = 1;                dp[u]|=dp[v];            }        }        for(int i=1; i<=m; i++){            int u = q[i].u;            int v = q[i].v;            for(int j=head[u]; ~j; j=E[j].next){                if(E[j].v!=v&&dp[E[j].v][v]){                    ans++;                    break;                }            }        }        io.wint(ans);    }    return 0;}

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