题解

6 条题解

  • 2
    @ 2015-05-03 19:32:08

    lct切之

    const
    inf = 1 << 30;
    type
    splayType = record
    rev : boolean;
    f,path_f,max,p : longint;
    c : array[0..1] of longint;
    end;
    eType = record
    x,y,a,b : longint;
    end;
    var
    e : array[0..100000] of eType;
    t : array[0..150000] of splayType;
    ans,n,m : longint;

    procedure swap(var x,y:longint);var z : longint; begin z := x; x := y; y := z; end;
    function min(x,y:longint):longint; begin min := x; if y < x then min := y; end;

    procedure reverse(x:longint);
    begin
    if x = 0 then exit;
    t[x].rev := not t[x].rev;
    swap(t[x].c[0],t[x].c[1]);
    end;

    procedure normalize(x:longint);
    var
    i : longint;
    begin
    if x = 0 then exit;
    if t[x].rev then
    begin
    t[x].rev := false;
    reverse(t[x].c[0]);
    reverse(t[x].c[1]);
    end;
    i := t[x].c[0];
    if t[t[x].c[1]].max > t[t[x].c[0]].max then i := t[x].c[1];
    if (x > n) and (t[i].max < e[x-n].b) then
    begin
    i := x; t[x].max := e[x-n].b; t[x].p := x;
    end;
    t[x].p := t[i].p; t[x].max := t[i].max;
    end;

    function dd(x:longint):longint;
    begin
    dd := ord(t[t[x].f].c[1] = x);
    end;

    procedure sc(x,y,d:longint);
    begin
    t[y].c[d] := x; t[x].f := y;
    end;

    procedure cc(x,d:longint);
    begin
    if t[x].c[d] = 0 then exit;
    t[t[x].c[d]].f := 0; t[t[x].c[d]].path_f := x; t[x].c[d] := 0;
    end;

    procedure rot(x:longint);
    var
    y,d : longint;
    begin
    y := t[x].f; d := dd(x);
    sc(x,t[y].f,dd(y));
    sc(t[x].c[d xor 1],y,d);
    sc(y,x,d xor 1);
    t[x].path_f := t[y].path_f;
    normalize(y);
    normalize(x);
    end;

    procedure splay(x,y:longint);
    begin
    while t[x].f <> y do
    begin
    normalize(t[t[x].f].f);
    normalize(t[x].f);
    normalize(x);
    if t[t[x].f].f <> y then
    if dd(t[x].f) = dd(x) then rot(t[x].f) else rot(x);
    rot(x);
    end;
    end;

    procedure access(x:longint);
    var
    y : longint;
    begin
    splay(x,0); normalize(x);
    cc(x,1); normalize(x);
    while t[x].path_f > 0 do
    begin
    y := t[x].path_f; splay(y,0);
    normalize(y); cc(y,1);
    sc(x,y,1); normalize(y);
    x := y;
    end;
    end;

    procedure evert(x:longint);
    begin
    access(x);
    splay(x,0);
    reverse(x);
    end;

    function exist(x,y:longint):boolean;
    begin
    exist := false;
    evert(x);
    access(y);
    splay(x,0);
    normalize(x);
    while y > 0 do
    begin
    if y = x then exit(true);
    y := t[y].f;
    end;
    end;

    function get(x,y:longint):longint;
    begin
    evert(x);
    access(y);
    splay(x,0);
    get := t[x].max;
    end;

    procedure cut(x,y:longint);
    begin
    evert(y);
    access(x);
    splay(y,0);
    t[y].c[1] := 0; t[x].f := 0; t[x].path_f := 0; normalize(y);
    end;

    procedure link(x,y:longint);
    begin
    evert(x);
    evert(y);
    splay(x,0);
    splay(y,0);
    normalize(x); sc(y,x,0); normalize(x);
    end;

    procedure sort(l,r:longint);
    var
    i,j : longint; x : eType;
    begin
    i := l; j := r; x := e[(l+r) >> 1];
    repeat
    while e[i].a < x.a do inc(i);
    while x.a < e[j].a do dec(j);
    if i <= j then
    begin
    e[0] := e[i]; e[i] := e[j]; e[j] := e[0];
    inc(i); dec(j);
    end;
    until i > j;
    if l < j then sort(l,j);
    if i < r then sort(i,r);
    end;

    procedure init;
    var
    j,i : longint;
    begin
    ans := inf;
    readln(n,m);
    for i := 1 to m do readln(e[i].x,e[i].y,e[i].a,e[i].b);
    sort(1,m);
    for i := 1 to m do t[i+n].max := e[i].b;
    for i := 1 to m do t[i+n].p := i+n;
    for i := 1 to m do
    begin
    if not exist(e[i].x,e[i].y) then
    begin
    link(e[i].x,i+n);
    link(e[i].y,i+n);
    end else begin
    if get(e[i].x,e[i].y) <= e[i].b then continue;
    j := t[e[i].x].p - n;
    cut(j+n,e[j].x); cut(j+n,e[j].y);
    link(e[i].x,i+n); link(e[i].y,i+n);
    end;
    if exist(1,n) then ans := min(ans,get(1,n)+e[i].a);
    end;
    if ans < inf then writeln(ans) else writeln(-1);
    end;

    begin
    init;
    end.

  • 0
    @ 2015-09-03 14:57:05

    P1865魔法森林
    Accepted

    记录信息

    评测状态 Accepted
    题目 P1865 魔法森林
    递交时间 2015-09-03 14:56:05
    代码语言 C++
    评测机 VijosEx
    消耗时间 11863 ms
    消耗内存 11180 KiB
    评测时间 2015-09-03 14:56:20

    评测结果

    编译成功

    foo.cpp: In function 'void spfa(int)':
    foo.cpp:58:24: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
    for( i = 0 ; i < linker[ now ].size() ; i++ )
    ^

    测试数据 #0: Accepted, time = 15 ms, mem = 4084 KiB, score = 5

    测试数据 #1: Accepted, time = 6 ms, mem = 4080 KiB, score = 5

    测试数据 #2: Accepted, time = 15 ms, mem = 4088 KiB, score = 5

    测试数据 #3: Accepted, time = 15 ms, mem = 4232 KiB, score = 5

    测试数据 #4: Accepted, time = 37 ms, mem = 4228 KiB, score = 5

    测试数据 #5: Accepted, time = 31 ms, mem = 4232 KiB, score = 5

    测试数据 #6: Accepted, time = 93 ms, mem = 4784 KiB, score = 5

    测试数据 #7: Accepted, time = 78 ms, mem = 4788 KiB, score = 5

    测试数据 #8: Accepted, time = 93 ms, mem = 4788 KiB, score = 5

    测试数据 #9: Accepted, time = 93 ms, mem = 4780 KiB, score = 5

    测试数据 #10: Accepted, time = 1312 ms, mem = 11128 KiB, score = 5

    测试数据 #11: Accepted, time = 1218 ms, mem = 11140 KiB, score = 5

    测试数据 #12: Accepted, time = 1046 ms, mem = 11132 KiB, score = 5

    测试数据 #13: Accepted, time = 1156 ms, mem = 11180 KiB, score = 5

    测试数据 #14: Accepted, time = 1375 ms, mem = 11132 KiB, score = 5

    测试数据 #15: Accepted, time = 1250 ms, mem = 11132 KiB, score = 5

    测试数据 #16: Accepted, time = 1031 ms, mem = 11140 KiB, score = 5

    测试数据 #17: Accepted, time = 828 ms, mem = 11132 KiB, score = 5

    测试数据 #18: Accepted, time = 1078 ms, mem = 11144 KiB, score = 5

    测试数据 #19: Accepted, time = 1093 ms, mem = 11128 KiB, score = 5

    Accepted, time = 11863 ms, mem = 11180 KiB, score = 100

    代码

    #include <iostream>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #include <string.h>
    #include <queue>

    using namespace std;

    int n , m;
    int i;
    vector < int > linker[50000 + 2];
    vector < int > di[50000 + 2];
    vector < int > fi[50000 + 2];

    struct edge
    {
    int x , y , a , b;
    };

    edge e[100000 + 2];

    inline int cmp( edge a , edge b )
    {
    return a.a < b.a;
    }

    bool use[50000 + 2];
    int dist[50000 + 2];
    int now , cur , v;

    inline int min( int a , int b )
    {
    if( a > b )
    return b;
    return a;
    }

    inline int max( int a , int b )
    {
    if( a > b )
    return a;
    return b;
    }

    int ans;
    queue < int > q;

    inline void spfa( int lim )
    {
    int i;
    q.push( 1 );
    while( !q.empty() )
    {
    now = q.front();
    q.pop();
    use[ now ] = 0;
    for( i = 0 ; i < linker[ now ].size() ; i++ )
    if( fi[ now ][i] <= lim )
    {
    cur = linker[ now ][i];
    v = di[ now ][i];
    if( dist[ cur ] > max( dist[ now ] , v ) )
    {
    dist[ cur ] = max( dist[ now ] , v );
    if( !use[ cur ] )
    {
    q.push( cur );
    use[ cur ] = 1;
    }
    }
    }
    }
    ans = min( ans , dist[n] + lim );
    return;
    }

    int main()
    {
    scanf( "%d %d" , &n , &m );
    for( i = 1 ; i <= m ; i++ )
    {
    scanf( "%d %d %d %d" , &e[i].x , &e[i].y , &e[i].a , &e[i].b );
    linker[ e[i].x ].push_back( e[i].y );
    linker[ e[i].y ].push_back( e[i].x );
    di[ e[i].x ].push_back( e[i].b );
    di[ e[i].y ].push_back( e[i].b );
    fi[ e[i].x ].push_back( e[i].a );
    fi[ e[i].y ].push_back( e[i].a );
    }
    sort( e + 1 , e + m + 1 , cmp );
    ans = 100000000;
    for( i = 2 ; i <= n ; i++ )
    dist[i] = 100000000;
    dist[1] = 0;
    for( i = 1 ; i <= m ; i++ )
    if( e[i].a >= ans )
    break;
    else
    {
    q.push( e[i].x );
    q.push( e[i].y );
    spfa( e[i].a );
    }
    if( ans <= 1000000 )
    cout << ans << endl;
    else
    cout << -1 << endl;
    return 0;
    }
    没有优化的spfa动态加点

  • 0
    @ 2015-05-22 09:38:17

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    #include<vector>
    #include<queue>
    #include<cstring>
    using namespace std;
    struct Tb{
    int num,aa,bb,next;
    };
    Tb a[200001];
    int first[100001];
    Tb p[100001];
    bool vis[100001];
    int n,m,ans;
    int sou(int v)
    {
    int j,k;
    if(v==n)
    {
    ans=min(ans,p[v].aa+p[v].bb);
    return 0;
    }
    k=first[v];
    while(a[k].num!=0)
    {
    j=a[k].num;
    if(!vis[j])
    {
    p[j].aa=max(p[v].aa,a[k].aa);
    p[j].bb=max(p[v].bb,a[k].bb);
    vis[j]=true;
    sou(j);
    vis[j]=false;
    }
    k=a[k].next;
    }

    }
    int main()
    {
    int w1,w2,x,y,k=0;
    freopen("p1865.in","r",stdin);
    scanf("%d%d",&n,&m);
    Tb b;
    for(int i=1;i<=m;i++)
    {
    scanf("%d%d%d%d",&x,&y,&w1,&w2);
    k++;
    a[k].next=first[x];
    a[k].num=y;
    a[k].aa=w1;
    a[k].bb=w2;
    first[x]=k;

    k++;
    a[k].next=first[y];
    a[k].num=x;
    a[k].aa=w1;
    a[k].bb=w2;
    first[y]=k;

    }
    memset(vis,false,true);
    ans=12345678;
    vis[1]=true;
    sou(1);
    if(ans==12345678)
    printf("-1");
    else printf("%d",ans);
    return 0;
    }

  • -1
    @ 2015-05-22 09:38:36

    hahaha

  • 1

信息

ID
1865
难度
5
分类
(无)
标签
递交数
257
已通过
81
通过率
32%
被复制
2
上传者