bzoj1924(强连通分量+各种STL运用)

来源：互联网发布：库管erp软件编辑：程序博客网时间：2024/06/01 16:16

在给定一些条件，自己建有向图，并在图中寻找经过点数最多的路径

：建图后缩点，记忆化搜索即可

#include<cstdio>#include<cstring>#include<stack>#include<cmath>#include<algorithm>#include<map>#include<cstdlib>#include<utility>#include<vector>#define fi first#define se second#define MK(a,b) make_pair((a),(b))#define pii pair<int,int>#define N 100005using namespace std;const int qx[]={0,0,1,1,1,-1,-1,-1};const int qy[]={1,-1,1,0,-1,1,0,-1};int n,r,c;map< pii,int> mp;vector< pii > xx[1000005],yy[1000005];bool pan(int x,int y) {if (x<1||y<1||x>r||y>c) return false;return mp.find(MK(x,y))!=mp.end();}struct aa{int x,y,t;}a[N];int val[N],pos[N],num;struct aaa{int head[N],tot,pre[N*50],to[N*50];void addedge(int u,int v) {to[++tot]=v;pre[tot]=head[u];head[u]=tot;}}old,now;void build(){for (int i=1;i<=r;i++) sort(xx[i].begin(),xx[i].end());for (int i=1;i<=c;i++) sort(yy[i].begin(),yy[i].end());int x,y,t;for (int i=1;i<=n;i++){x=a[i].x,y=a[i].y,t=a[i].t;if (t==1){int size=xx[x].size();for (int j=0;j<size;j++) if(xx[x][j].fi!=y) old.addedge(i,xx[x][j].se);}if (t==2){int size=yy[y].size();for (int j=0;j<size;j++) if (yy[y][j].fi!=x) old.addedge(i,yy[y][j].se); }if (t==3){int xt,yt;for (int j=0;j<8;j++)if (pan(xt=x+qx[j],yt=y+qy[j])) old.addedge(i,mp[MK(xt,yt)]);}}}int dfn[N],low[N],cnt;bool in[N];stack<int> s;void dfs(int u){low[u]=dfn[u]=++cnt;in[u]=true;s.push(u);for (int i=old.head[u];i;i=old.pre[i]){int v=old.to[i];if (!dfn[v]){dfs(v);low[u]=min(low[u],low[v]);}else if (in[v]) low[u]=min(low[u],dfn[v]);}if (low[u]==dfn[u]){int v;num++;do{v=s.top();s.pop();pos[v]=num;in[v]=false;val[num]++;}while (v!=u);}}void rebuild(){for (int u=1;u<=n;u++)for (int v,i=old.head[u];i;i=old.pre[i]){v=old.to[i];if (pos[u]!=pos[v]) now.addedge(pos[u],pos[v]);}}int f[N];int dp(int u){if (f[u]!=-1) return f[u];int v;f[u]=0;for (int i=now.head[u];i;i=now.pre[i]) f[u]=max(f[u],dp(now.to[i]));f[u]+=val[u];return f[u];}int main(){scanf("%d%d%d",&n,&r,&c);int x,y,t;for (int i=1;i<=n;i++){scanf("%d%d%d",&x,&y,&t);a[i]=(aa){x,y,t};mp[MK(x,y)]=i;xx[x].push_back(MK(y,i));yy[y].push_back(MK(x,i));}build();for (int i=1;i<=n;i++) if (!dfn[i]) dfs(i);rebuild();memset(f,-1,sizeof(f));int ans=0;for (int i=1;i<=num;i++) ans=max(dp(i),ans);printf("%d",ans);return 0;}

总结

1：一般用到缩点的题目的步骤

读入原图old——强连通分量缩点——缩点后直接建新图——在新图上处理问题

2：stl，vector处理行列，map处理八连通的大大减少了编程复杂度。

基本的stl必须练熟

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