摘要：一开始我一直想的是简单图论问题，构图SPFA求最长路，发现这样做有环不好处理。比如这组样例:21 1 1 01 1 1 0构图就会有环，不能直接SPFA，tarjan缩点的话应该可以做，但是代码量就太大了。可以这样去处理：bool cmp(block a,block b) { if(a.x!=b.x)return a.x<b.x; if(a.y!=b.y)return a.y<b.y; return a.d>b.d; } void build(){ //缩点，避免产生环 sort(x,x+n,cmp); int cnt=1; y[1]=x[0]; for(int i=1;i& 阅读全文

摘要：MABODXAccept:17 Submit:88Time Limit:4000MS Memory Limit:65536KBDescriptionThere is a country which has N city. Because The king spend littlemoney on the road ,so there is one and only one path from any cityto another city,every city has a apple tree.The apple in one citywhich has only one apple tree 阅读全文

摘要：2-SAT经典问题#include<cstdio> #include<string.h> #include<math.h> #include<stack> #define N 2010 #define M N*N*3 using namespace std; struct edge{ int v,next; }edge[M]; int n; int s[N],t[N],seq[N]; int cnt,head1[N],head2[N],head3[N]; int scc,index,dfn[N],low[N],belong[N]; int top 阅读全文

摘要：很简单的一道题，先tarjan缩点，然后SPFA求最长路。#include<cstdio> #include<string.h> #include<math.h> #include<queue> #define N 30100 #define M 150100 using namespace std; struct edge{ int v,next; }edge[M]; int n,m; int num[N]; int cnt,head1[N],head2[N]; int scc,index,dfn[N],low[N],belong[N],sum 阅读全文

摘要：这个题算是LCA的一个高级应用吧。其实就是在维护u,v到其最近公共祖先的几个数据（将u,v分成两段,一个从u到LCA(u,v),一个从LCA(u,v)到v,分别维护即可）#include<cstdio> #include<string.h> #include<math.h> #define con 50100 using namespace std; int head1[con],head2[con],head3[con],p,n,num[con],mi[con],mx[con],up[con],down[con],vis[con],f[50100],ans 阅读全文

摘要：Tarjan算法（离线算法）模板#include<cstdio> #include<string.h> using namespace std; int n,head[10010],k,x,y,d[10010],f[10010],ans[10010],num[10010]; bool visit[10010]; struct point{ int to; int next; }edge[10010]; void addedge(int a,int b){ edge[k].to=b; edge[k].next=head[a]; head[a]=k++; } int fin 阅读全文