The first chapter of konjac solution to a problem, bigwigs light spray (shivering
Closer to home, this question
see the solution to a problem area bigwigs with a wide variety of algorithms gods cut out the problem, and does not indicate a small konjac algorithm so many gods, and thus be out of the water solution to a problem to talk about their own practices.
Algorithm: Tarjan + topological sorting + DP
Ideas: see this question first thought tarjan shrink point (I will not tell you this is because I will only tarjanAfter tarjan a directed acyclic graph, naturally conceivable dp topologically sorted, then the general idea of solving the problem came out.
The first step tarjan point reduction
void tarjan(int u){
dfn[u]=low[u]=++num;
s[++temp]=u;
for(int i=head[u];i;i=e[i].next){
int v=e[i].to;
if(!dfn[v]){
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(!color[v]) low[u]=min(low[u],low[v]);
}
if(dfn[u]==low[u]){
color[u]=++sum;
ma[sum]=max(ma[sum],cost[u]);//维护强连通分量中最大的商品价格
mi[sum]=min(mi[sum],cost[u]);//维护强连通分量中最小的商品价格
while(s[temp]!=u){
ma[sum]=max(ma[sum],cost[s[temp]]);
mi[sum]=min(mi[sum],cost[s[temp]]);//同上
color[s[temp--]]=sum;
}
--temp;
}
}Step topological sorting + DP
State transition equation dp's quite easy to think of it.
dp[to]=max{ma[to]-mi[k]}
To here represents the point to be found, mi [k] represents the minimum value of the entry path commodity prices
inline void tuopu(){
queue<int> q;
q.push(color[1]);
ans[color[1]]=ma[color[1]]-mi[color[1]];
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=head[u];i;i=ne[i].next){
int v=ne[i].to;
--rd[v];
mi[v]=min(mi[v],mi[u]);//维护路径上经过的最小值
ans[v]=max(ans[u],ma[v]-mi[v]);//dp状态转移
if(!rd[v]) q.push(v);
}
}
}Here's the key code for this problem will be solved
The following release total code
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define INF 2e9+11;
using namespace std;
const int N=100000+11;
const int M=500000+11;
int n,m,sum,num,temp,tot;
int head[N],dfn[N],low[N],s[N],color[N],ma[N],mi[N],cost[N],rd[N],ans[N];
struct Edge{
int from,next,to;
}e[M],ne[M];
inline void read(int &a){
a=0;
char c=getchar();
while(c>57 or c<48)c=getchar();
while(47<c and c<58){
a=a*10+c-48;
c=getchar();
}
}
inline int max(int x,int y){return x > y ? x : y;}
inline int min(int x,int y){return x > y ? y : x;}
inline void add_edge(int from,int to){
e[++tot].next=head[from];
e[tot].from=from;
e[tot].to=to;
head[from]=tot;
}//链式前向星建边
void tarjan(int u){
dfn[u]=low[u]=++num;
s[++temp]=u;
for(int i=head[u];i;i=e[i].next){
int v=e[i].to;
if(!dfn[v]){
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(!color[v]) low[u]=min(low[u],low[v]);
}
if(dfn[u]==low[u]){
color[u]=++sum;
ma[sum]=max(ma[sum],cost[u]);
mi[sum]=min(mi[sum],cost[u]);
while(s[temp]!=u){
ma[sum]=max(ma[sum],cost[s[temp]]);//维护强连通分量中最大的商品价格
mi[sum]=min(mi[sum],cost[s[temp]]);//维护强连通分量中最小的商品价格
color[s[temp--]]=sum;
}
--temp;
}
}
inline void tuopu(){
queue<int> q;
q.push(color[1]);
ans[color[1]]=ma[color[1]]-mi[color[1]];
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=head[u];i;i=ne[i].next){
int v=ne[i].to;
--rd[v];
mi[v]=min(mi[v],mi[u]);//维护路径上经过的最小值
ans[v]=max(ans[u],ma[v]-mi[v]);//dp状态转移
if(!rd[v]) q.push(v);
}
}
}
int main(){
read(n);read(m);
for(int i=1;i<=n;++i) read(cost[i]);
int x,y,z;
for(int i=1;i<=m;++i){
read(x);read(y);read(z);
if(z&1) add_edge(x,y);
else{
add_edge(x,y);
add_edge(y,x);
}
}
for(int i=1;i<=n;++i) mi[i]=INF;//初始化mi数组
for(int i=1;i<=n;++i)
if(!dfn[i]) tarjan(i);
tot=0;
memset(head,0,sizeof(head));
for(int i=1;i<=m;++i)
if(color[e[i].from]!=color[e[i].to]){
++rd[color[e[i].to]];
ne[++tot].next=head[color[e[i].from]];
ne[tot].to=color[e[i].to];
head[color[e[i].from]]=tot;
}//缩点之后的新图建边
tuopu();
printf("%d",ans[color[n]]);
return 0;
}If you do not know where I can private letter, if there is an error where chiefs welcome to correct me (Welcome to beating and hanging my deities