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zc
2019-12-31 11:31:00
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先想想暴力做法:

bfs出不涉水可以到达的点,然后在这些点中找出与点1的最小距离

优化:

我们可以使用kruskal重构树来快速求出这些点

先按海拔从高到低排序,这样见出来的kruskal重构树的是海拔的小根堆

我们可以倍增找出最远可以到达的祖先,然后求出这段区间中的点与点1的最小距离

可以在dfs的时候预处理

详见代码

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 200011
int n,m,cnt=0,d[N<<1],head[N];
struct link{
    int x,y,l,a;
    il bool operator<(link b){return a>b.a;}
}E[N<<1];
struct edge{int to,nxt,w;}e[N<<2];
il void add(int x,int y,int w){e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;e[cnt].w=w;}
bool v[N];
il void dij(){
    struct cmp{il bool operator()(int x,int y){return d[x]>d[y];}};
    priority_queue<int,vector<int>,cmp>q;
    Fur(i,1,n)d[i]=inf;
    d[1]=0;
    q.push(1);
    while(!q.empty()){
        int x=q.top();v[x]=0;q.pop();
        fl(i,x)if(d[x]+e[i].w<d[to]){
            d[to]=d[x]+e[i].w;
            if(!v[to])v[to]=1,q.push(to);
        }
    }
}
int fa[N<<1],ne[N<<1],f[21][N<<1];
int gf(int x){return (x==fa[x])?x:(fa[x]=gf(fa[x]));}
int ls[N<<1],rs[N<<1];
void dfs(int x){
    Fur(i,1,20)f[i][x]=f[i-1][f[i-1][x]];
    if(!ls[x])return;
    dfs(ls[x]),dfs(rs[x]);
    d[x]=MIN(d[ls[x]],d[rs[x]]);
}
il void work(){
    clr(head,0);cnt=0;
    clr(ls,0);clr(rs,0);
    in(n,m);
    int x,y,l,a,t=n;
    Fur(i,1,m){
        in(x,y,l,a);
        E[i]=link{x,y,l,a};
        add(x,y,l);add(y,x,l);
    }
    dij();
    sort(E+1,E+m+1);
    Fur(i,1,n*2+1)fa[i]=i;
    Fur(i,1,m){
        x=gf(E[i].x),y=gf(E[i].y);
        if(x==y)continue;
        ne[++t]=E[i].a;
        fa[x]=fa[y]=fa[t]=t;
        f[0][x]=f[0][y]=t;
        ls[t]=x;rs[t]=y;
    }
    dfs(t);
    int Q,K,S,p,la=0;
    in(Q,K,S);
    while(Q--){
        in(x,p);
        x=(x+K*la-1)%n+1;
        p=(p+K*la)%(S+1);
        Fdr(i,20,0)if(f[i][x]&&ne[f[i][x]]>p)x=f[i][x];
        out(la=d[x],ln);
    }
}
int main(){
    int T;in(T);
    while(T--)work();
    flush();
}
LG 4768 [NOI2018]归程
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