Codeforces IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2)

来源：互联网发布：mysql disable索引编辑：程序博客网时间：2024/06/06 01:38

C. Bear and Up-Down

当不满足要求的数超过4个的时候，肯定是无解的，否则用每个数和这4个数交换，然后check。。比赛的时候写了很繁琐的分类讨论，竟然没挂。。

#include <bits/stdc++.h>#include <unordered_map>#define ll long long using namespace std;int a[150010];int err[150010];int n;bool judge(){    for(int i=1;i<n;i++){        if(i&1){            if(a[i]>=a[i+1]){                return 0;            }        }else{            if(a[i]<=a[i+1]){                return 0;            }        }    }    return 1;}int main(){    cin>>n;    for(int i=1;i<=n;i++){        scanf("%d",&a[i]);    }    int cnt= 0;    for(int i=1;i<n;i++){        if(i&1){            if(a[i]>=a[i+1]){                err[cnt++] = i;            }        }else{            if(a[i]<=a[i+1]){                err[cnt++] = i;            }        }    }    int ans = 0;    if(cnt>4){    }else if(cnt==4){        if(err[0]+1==err[1] && err[2]+1==err[3]){            swap(a[err[1]],a[err[3]]);            ans+=judge();            swap(a[err[1]],a[err[3]]);        }else{        }    }else if(cnt==3){        if(err[0]+2==err[2]){            swap(a[err[1]],a[err[2]]);            ans+=judge();            swap(a[err[1]],a[err[2]]);            swap(a[err[1]],a[err[2]+1]);            ans+=judge();            swap(a[err[1]],a[err[2]+1]);            swap(a[err[0]],a[err[2]]);            ans+=judge();            swap(a[err[0]],a[err[2]]);        }if(err[0]+1==err[1]){            swap(a[err[1]],a[err[2]]);            ans+=judge();            swap(a[err[1]],a[err[2]]);            swap(a[err[1]],a[err[2]+1]);            ans+=judge();            swap(a[err[1]],a[err[2]+1]);        }else if(err[1]+1==err[2]){            swap(a[err[0]],a[err[2]]);            ans+=judge();            swap(a[err[0]],a[err[2]]);            swap(a[err[0]+1],a[err[2]]);            ans+=judge();            swap(a[err[0]+1],a[err[2]]);        }else{        }    }else if(cnt==2){        if(err[0]+1==err[1]){            swap(a[err[0]],a[err[1]+1]);            ans+=judge();            swap(a[err[0]],a[err[1]+1]);            for(int i=1;i<=n;i++){                swap(a[i],a[err[1]]);                bool ok = 1;                for(int j=i-3;j<=i+3;j++){                    if(j<1)continue;                    if(j>=n)continue;                    if(j&1){                        if(a[j]>=a[j+1]){                            ok = 0;                        }                    }else{                        if(a[j]<=a[j+1]){                            ok = 0;                        }                    }                }                for(int j=err[1]-3;j<=err[1]+3;j++){                    if(j<1)continue;                    if(j>=n)continue;                    if(j&1){                        if(a[j]>=a[j+1]){                            ok = 0;                        }                    }else{                        if(a[j]<=a[j+1]){                            ok = 0;                        }                    }                }                swap(a[i],a[err[1]]);                ans+=ok;            }        }else{            swap(a[err[0]],a[err[1]]);            ans+=judge();            swap(a[err[0]],a[err[1]]);            swap(a[err[0]],a[err[1]+1]);            ans+=judge();            swap(a[err[0]],a[err[1]+1]);            swap(a[err[0]+1],a[err[1]]);            ans+=judge();            swap(a[err[0]+1],a[err[1]]);            swap(a[err[0]+1],a[err[1]+1]);            ans+=judge();            swap(a[err[0]+1],a[err[1]+1]);        }    }else if(cnt==1){        for(int i=1;i<=n;i++){            swap(a[i],a[err[0]]);            bool ok = 1;            for(int j=i-3;j<=i+3;j++){                if(j<1)continue;                if(j>=n)continue;                if(j&1){                    if(a[j]>=a[j+1]){                        ok = 0;                    }                }else{                    if(a[j]<=a[j+1]){                        ok = 0;                    }                }            }            for(int j=err[0]-3;j<=err[0]+3;j++){                if(j<1)continue;                if(j>=n)continue;                if(j&1){                    if(a[j]>=a[j+1]){                        ok = 0;                    }                }else{                    if(a[j]<=a[j+1]){                        ok = 0;                    }                }            }            swap(a[i],a[err[0]]);            ans+=ok;        }        for(int i=1;i<=n;i++){            if(i==err[0])continue;            swap(a[i],a[err[0]+1]);            bool ok = 1;            for(int j=i-3;j<=i+3;j++){                if(j<1)continue;                if(j>=n)continue;                if(j&1){                    if(a[j]>=a[j+1]){                        ok = 0;                    }                }else{                    if(a[j]<=a[j+1]){                        ok = 0;                    }                }            }            for(int j=err[0]-3;j<=err[0]+3;j++){                if(j<1)continue;                if(j>=n)continue;                if(j&1){                    if(a[j]>=a[j+1]){                        ok = 0;                    }                }else{                    if(a[j]<=a[j+1]){                        ok = 0;                    }                }            }            swap(a[i],a[err[0]+1]);            ans+=ok;        }    }    cout<<ans<<endl;    return 0;}

D. Delivery Bears

这是一道好题！解法是二分答案，然后跑最大流判是否可行。判断方法是将边的流量改为它除以答案的值，向下取整，如果最大流不小于x就是合法。
另外要注意，二分写成循环固定次数会比较好。

#include <bits/stdc++.h>#include <unordered_map>#define ll long long using namespace std;int n,m;double x;const int inf=0xfffff;  const int N = 55;  int  pre[N];double maxflow,dp[N][N];double tmp[N][N];void Edmonds_Karp(int start, int end, int m){      while(1){          queue<int> p;          double minflow = inf;          p.push(1);        memset(pre, 0, sizeof(pre));        while(!p.empty()){            int u = p.front();              p.pop();              if(u == end)                               break;              for(int i = 1;i <= m;i++)                  if(dp[u][i] > 0&&pre[i] == 0){                       pre[i] = u;                                    p.push(i);                  }              }          if(pre[end] == 0)              break;          for(int i = end;i != start;i = pre[i])              minflow = min(minflow, dp[pre[i]][i]);          for(int i = end;i != start;i = pre[i])  {              dp[pre[i]][i] -= minflow;             dp[i][pre[i]] += minflow;          }          maxflow+=minflow;      }  } int main(){    cin>>n>>m>>x;    for(int i=1;i<=m;i++){        int a,b,c;        scanf("%d%d%d",&a,&b,&c);        dp[a][b]+=c;    }    memcpy(tmp,dp,sizeof(dp));    double l=0;    double r=1000000;    double mid = 0;    //注意这个二分的姿势     for(int tt=0;tt<66;tt++){        mid = (l+r)/2;        memcpy(dp,tmp,sizeof(dp));        for(int i=0;i<=n;i++){            for(int j=0;j<=n;j++){                dp[i][j] = floor(dp[i][j]/mid);            }        }        maxflow = 0;        Edmonds_Karp(1,n,n);        if(maxflow>=x){            l=mid;        }else{            r=mid;        }    }    printf("%.10f\n",mid*x);    return 0;}

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