【SDOI2015】【BZOJ4086】旅行计划travel

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本文针对SDOI2015旅行问题提供了一种详细的算法解决方案，该问题涉及复杂的路径寻找与状态更新。文章通过不同的k值分别设计了特定的算法，并运用容斥原理来优化解题策略。

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4086: [Sdoi2015]travel

Time Limit: 80 Sec Memory Limit: 512 MB
Submit: 54 Solved: 27
[Submit][Status][Discuss]
Description

Input

Output

Sample Input

Sample Output

HINT

Source
题面去Vijos看.
对每种k需要单独设计一个算法….
主要是要用考虑对方案进行容斥…
感觉doc老师博客里给的标算其实本质也是容斥?
因为为了压代码牺牲了代码的常数,导致跑的特别慢呢..

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#define MAXN 1010
#define MAXM 10010
#define GET (ch>='0'&&ch<='9')
using namespace std;
inline void in(int &x)
{
    char ch=getchar();x=0;
    while (!GET)    ch=getchar();
    while (GET) x=x*10+ch-'0',ch=getchar();
}
int T,n,m,k,top,sum,tot;
int a[MAXN][MAXN],l[MAXN][MAXN],cnt[MAXN],num[MAXN][MAXN],mat[MAXN][MAXN];
int c1[MAXN][MAXN],c2[MAXN][MAXN];
vector<int> c[MAXN][MAXN];
struct edge {   int to; edge *next; }e[MAXM],*prev[MAXN];
inline void insert(int u,int v) {   e[++top].to=v;e[top].next=prev[u];prev[u]=&e[top];  }
inline void solve_2()
{
    for (register int i=1;i<=n;++i)  for (register int j=1;j<=n;++j)  l[i][j]|=a[i][j];
}
inline void solve_3()
{
    for (register int x=1;x<=n;x++)
    for (register edge *i=prev[x];i;i=i->next)   for (register edge *j=prev[i->to];j;j=j->next)
        if (j->to!=x)    l[x][j->to]=1;
}
inline void solve_4()
{
    for (register int i=1;i<=n;++i)  for (register int j=i+1;j<=n;++j)
        for (register edge *I=prev[i];I;I=I->next)
            if (I->to!=j)
                for (register edge *J=prev[j];J;J=J->next)
                    if (J->to!=I->to&&J->to!=i)    l[i][j]=(l[j][i]|=a[I->to][J->to]);
}
inline void solve_5()
{
    for (register int i=1;i<=n;++i)  for (register int j=i+1;j<=n;++j)
    {
        for (int k=1;k<=n;++k)   cnt[k]=0;sum=0;
        for (register edge *I=prev[i];I;I=I->next)   if (I->to!=j&&a[I->to][j])    ++cnt[I->to],++sum;
        for (register edge *I=prev[i];I;I=I->next)
            if (I->to!=j)
                for (register edge *J=prev[j];J;J=J->next)
                    if (J->to!=i&&J->to!=I->to)    l[I->to][J->to]=(l[J->to][I->to]|=(sum-cnt[I->to]-cnt[J->to]>0));
    }
}
inline void solve_6()
{
    for (register int i=1;i<=n;++i)  for (register int j=i+1;j<=n;++j)
    {
        for (register int k=1;k<=n;++k)  cnt[k]=0;sum=0;
        for (register edge *I=prev[i];I;I=I->next)
            if (I->to!=j)
                for (register edge *J=prev[j];J;J=J->next)   if (J->to!=i&&I->to!=J->to&&a[I->to][J->to]) ++cnt[I->to],++cnt[J->to],++sum;
        for (register edge *I=prev[i];I;I=I->next)
            if (I->to!=j)
                for (register edge *J=prev[j];J;J=J->next)
                    if (J->to!=i&&I->to!=J->to)    l[I->to][J->to]=(l[J->to][I->to]|=(sum-cnt[I->to]-cnt[J->to]+a[I->to][J->to]>0));
    }
}
inline void solve_7()
{
    for (register int i=1;i<=n;++i)  for (register int j=1;j<=n;++j)  c[i][j].clear(),num[i][j]=mat[i][j]=0;
    for (register int x=1;x<=n;++x)
        for (register edge *i=prev[x];i;i=i->next)   for (edge *j=prev[i->to];j;j=j->next)
            if (j->to!=x)    c[x][j->to].push_back(i->to);
    for (register int i=1;i<=n;++i)  for (register int j=i+1;j<=n;++j)
    {
        for (register int k=1;k<=n;++k)  cnt[k]=0;++tot;sum=0;
        for (register edge *I=prev[i];I;I=I->next)
        if (I->to!=j)
            for (register edge *J=prev[j];J;J=J->next)
            if (J->to!=i&&J->to!=I->to)
                for (register int k=c[I->to][J->to].size()-1;~k;k--)
                {
                    int t=c[I->to][J->to][k];
                    if (t==i||t==j) continue;
                    if (num[I->to][J->to]==tot)   ++c1[I->to][J->to];
                    else    c1[I->to][J->to]=1,num[I->to][J->to]=tot;
                    if (mat[I->to][t]==tot)  ++c2[I->to][t];
                    else    c2[I->to][t]=1,mat[I->to][t]=tot;
                    if (mat[J->to][t]==tot)  ++c2[J->to][t];
                    else    c2[J->to][t]=1,mat[J->to][t]=tot;
                    ++cnt[I->to];++cnt[J->to];++cnt[t];++sum;

                }
        for (register edge *I=prev[i];I;I=I->next)
        if (I->to!=j)
            for (register edge *J=prev[j];J;J=J->next)
            if (J->to!=I->to&&J->to!=i)    l[I->to][J->to]=(l[J->to][I->to]|=(sum-cnt[I->to]-cnt[J->to]+(num[I->to][J->to]==tot?c1[I->to][J->to]:0)+(mat[I->to][J->to]==tot?c2[I->to][J->to]:0)+(mat[J->to][I->to]==tot?c2[J->to][I->to]:0)>0));
    }
}
int main()
{
    for (in(T);T;T--)
    {
        in(n);in(m);in(k);int u,v;top=0;
        for (register int i=1;i<=n;++i)  prev[i]=0x0;
        for (register int i=1;i<=n;++i)  for (int j=1;j<=n;++j)   a[i][j]=l[i][j]=0;
        for (register int i=1;i<=m;++i)  in(u),in(v),insert(u,v),insert(v,u),a[u][v]=a[v][u]=1;
        if (k==2)   solve_2();
        if (k==3)   solve_3();
        if (k==4)   solve_4();
        if (k==5)   solve_5();
        if (k==6)   solve_6();
        if (k==7)   solve_7();
        for (register int i=1;i<=n;++i)
        {
            for (register int j=1;j<=n;++j)  putchar(l[i][j]?'Y':'N');
            putchar('\n');
        }
    }
}