二分解單調函數

POJ 3273

二分最小值

#include <stdio.h>
#include <iostream>
#include <queue>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#include <fstream>
#include <set>
#include <stack>
#include <list>
using namespace std;

#define READ freopen("acm.in","r",stdin)
#define WRITE freopen("acm.out","w",stdout)
#define ll long long
#define ull unsigned long long 
#define uint unsigned int
#define PII pair<int,int>
#define PDD pair<double,double>
#define fst first
#define sec second
#define MS(x,d) memset(x,d,sizeof(x))
#define INF 1000001000
#define ALL(x) x.begin(),x.end()
#define PB push_back
#define MOD 99991
#define MAX 111111


int n,m;
int mon[200000];

bool C(int x)
{
    int cnt=0;
    int sum=0;
    for(int i=0;i<n;i++)
    {
        if(mon[i]>x)
            return false;
        if(sum+mon[i]<=x)
            sum+=mon[i];
        else
            sum=mon[i],cnt++;
        if(cnt>m)
            return false;
    }
    cnt++;
    return cnt<=m; 
}

int solve()
{
    int lb=0,ub=INF;
    while(ub-lb>1)
    {
        //cout<<lb<<" "<<ub<<endl;
        int m=(ub+lb)/2;
        if(C(m))
            ub=m;
        else
            lb=m;
    }
    return ub;
}


int main()
{
    READ;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&mon[i]);
        cout<<solve()<<endl;
    }
    return 0;
}

POJ 3258

二分最大值

#include <stdio.h>
#include <iostream>
#include <queue>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#include <fstream>
#include <set>
#include <stack>
#include <list>
using namespace std;

#define READ freopen("acm.in","r",stdin)
#define WRITE freopen("acm.out","w",stdout)
#define ll long long
#define ull unsigned long long 
#define uint unsigned int
#define PII pair<int,int>
#define PDD pair<double,double>
#define fst first
#define sec second
#define MS(x,d) memset(x,d,sizeof(x))
#define INF 1000001000
#define ALL(x) x.begin(),x.end()
#define PB push_back
#define MOD 99991
#define MAX 111111


int L,n,m;
int len[200000];


bool C(int x)
{
    int l=0;// 上一個石頭和上上一個石頭之間的距離 如果上一個石頭沒被去掉則爲0
    int cnt=0;
    if(x>L)
        return false;
    for(int i=1;i<n;i++)
    {
        if(l+len[i]-len[i-1]<x)
            cnt++,l+=len[i]-len[i-1];
        else
            l=0;
    }
    return cnt<=m;
}

int solve()
{
    int lb=0,ub=INF;
    while(ub-lb>1)
    {
        int m=(lb+ub)>>1;
        if(C(m))
            lb=m;
        else
            ub=m;
    }
    return lb;
}


int main()
{
    READ;
    while(scanf("%d%d%d",&L,&n,&m)!=EOF)
    {
        len[0]=0;
        len[n+1]=L;
        for(int i=1;i<=n;i++)
            scanf("%d",&len[i]);
        n++;
        sort(len,len+n+1);
        cout<<solve()<<endl;
    }
    return 0;
}

 OJ  上的一道二分+tarjan

http://ecnu.acmclub.com/index.php?app=problem_title&id=29&problem_id=21730

#include <stdio.h>
#include <iostream>
#include <queue>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#include <fstream>
#include <set>
#include <stack>
#include <list>
using namespace std;

#define READ freopen("acm.in","r",stdin)
#define WRITE freopen("acm.out","w",stdout)
#define ll long long
#define ull unsigned long long 
#define uint unsigned int
#define PII pair<int,int>
#define PDD pair<double,double>
#define fst first
#define sec second
#define MS(x,d) memset(x,d,sizeof(x))
#define INF 1000001000
#define ALL(x) x.begin(),x.end()
#define PB push_back
#define MOD 99991
#define MAX 11111


int dfn[MAX],low[MAX];
bool instack[MAX];
stack<int> S;
int cnt,cnt_scc;
vector<int> G[MAX];
int n,m;
PII rel[200000];

int tarjan(int u)
{ 
    if(dfn[u]==-1)
    {
        dfn[u]=low[u]=++cnt;
        S.push(u),instack[u]=1;
        for(int i=0;i<G[u].size();i++)
        {
            int v=G[u][i];
            if(dfn[v]==-1)
                low[u]=min(low[u],tarjan(v));
            else if(instack[v])
                low[u]=min(dfn[v],low[u]);
        }
        if(dfn[u]==low[u])
        {
            while(instack[u])
            {
                instack[S.top()]=0;
                S.pop();
            }
            cnt_scc++;
        }
    }
    return low[u];
}

bool C(int x)
{
    MS(dfn,-1),cnt=cnt_scc=0,MS(instack,0);
    for(int i=0;i<MAX;i++)
        G[i].clear();
    while(!S.empty()) S.pop();
    for(int i=0;i<x;i++)
    {
        int f,t;
        f=rel[i].fst;
        t=rel[i].sec;
        G[f].PB(t);
    }
    for(int i=1;i<=n;i++)
    {
        if(dfn[i]==-1)
            tarjan(i);
    }
    return cnt_scc==n;
}

int solve()
{
    int lb=0,ub=m+1;
    while(ub-lb>1)
    {
        int mid=(ub+lb)>>1;
        if(C(mid))
            lb=mid;
        else
            ub=mid;
    }
    return lb;
}

int main()
{
   // READ;
    int cas;
    scanf("%d",&cas);
    for(int T=1;T<=cas;T++)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
            scanf("%d%d",&rel[i].fst,&rel[i].sec);
        printf("Case %d: %d\n",T,solve());
    }
    return 0;
}

二分總還是蠻。。。。厲害的。

這也是一種枚舉方式嘛，要求最大值最小值，基本就是二分的標誌了。

專心複習考試去了。。