Codevs1992题解

• 题目大意
  求有向图中经过某一点k的最大环（数据规模不支持floyd）。

• 题解
  以k为起点在正向图中spfa求单源最短路，再在反向图中spfa求单源最短路。枚举除k外的每一个点i，如果有一个同时包含i与k的环，ans=max{ans,dist[i]+invdist[i]}。

• Code

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int maxn = 1010, maxm = 200010, nil = 0, oo = 1061109567;int n, m, k;int pnt[maxn], nxt[maxm], u[maxm], v[maxm], w[maxm], e;int d[maxn], invd[maxn];bool vis[maxn], other[maxm];void addedge(int x, int y, int z){    u[++e] = x; v[e] = y; w[e] = z;    nxt[e] = pnt[x]; pnt[x] = e; other[e] = false;    u[++e] = y; v[e] = x; w[e] = z;    nxt[e] = pnt[y]; pnt[y] = e; other[e] = true;}void init(){    int x, y, z;    scanf("%d%d%d", &n, &m, &k);    for(int i = 1; i <= m; ++i)    {        scanf("%d%d%d", &x, &y, &z);        addedge(x, y, z);    }}void work(){    queue <int> q;    //以下为反向spfa    memset(invd, 0x3f, sizeof(invd));    memset(vis, 0, sizeof(vis));    invd[k] = 0; vis[k] = true;    q.push(k);    while(!q.empty())    {        int tmp = q.front();        q.pop();        vis[tmp] = false;        for(int j = pnt[tmp]; j != nil; j = nxt[j])        {            if(other[j] && invd[v[j]] > invd[tmp] + w[j])            {                invd[v[j]] = invd[tmp] + w[j];                vis[v[j]] = true;                q.push(v[j]);            }        }    }    //以下为正向spfa    memset(d, 0x3f, sizeof(d));    memset(vis, 0, sizeof(vis));    d[k] = 0; vis[k] = true;    q.push(k);    while(!q.empty())    {        int tmp = q.front();        q.pop();        vis[tmp] = false;        for(int j = pnt[tmp]; j != nil; j = nxt[j])        {            if((!other[j]) && d[v[j]] > d[tmp] + w[j])            {                d[v[j]] = d[tmp] + w[j];                vis[v[j]] = true;                q.push(v[j]);            }        }    }    int ans = 0;    for(int i = 1; i <= n; ++i)    {        if(d[i] != oo && invd[i] != oo)        {            ans = max(ans, d[i] + invd[i]);        }    }    printf("%d\n", ans);}int main(){    init();    work();    return 0;}