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  • Poj 1201 Intervals(差分约束)

    Intervals
    Time Limit: 2000MS Memory Limit: 65536K
    Description
    You are given n closed, integer intervals [ai, bi] and n integers c1, …, cn.
    Write a program that:
    reads the number of intervals, their end points and integers c1, …, cn from the standard input,
    computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,…,n,
    writes the answer to the standard output.
    Input
    The first line of the input contains an integer n (1 <= n <= 50000) – the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.
    Output
    The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,…,n.
    Sample Input
    5
    3 7 3
    8 10 3
    6 8 1
    1 3 1
    10 11 1
    Sample Output
    6
    Source
    Southwestern Europe 2002

    /*
    又回过头来看了一下半年前学的差分约束.
    感觉自己还是弱弱的. 
    由约束条件可得
    (1)dis[y+1]-dis[x]>=z.
    (2)0<=dis[i]-dis[i-1]<=1.
    因为是跑最长路.
    所以要把(2)式拆成
    dis[i]-dis[i-1]>=0.
    dis[i-1]-dis[i]>=-1.
    spfa松弛即可.
    */
    #include<cstring>
    #include<cstdio>
    #include<queue>
    #define MAXN 50001
    using namespace std;
    struct data{int v,next,x;}e[MAXN*3];
    int n,m,head[MAXN],dis[MAXN],cut,maxv,maxn,minn=1e6;
    bool b[MAXN];
    void add(int u,int v,int x)
    {
        e[++cut].v=v;
        e[cut].x=x;
        e[cut].next=head[u];
        head[u]=cut;
    }
    int read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
        return x*f;
    }
    void spfa()
    {
        memset(dis,-127/3,sizeof dis);
        queue<int>q;q.push(minn);dis[minn]=0;
        while(!q.empty())
        {
            int u=q.front();q.pop();b[u]=false;
            for(int i=head[u];i;i=e[i].next)
            {
                int v=e[i].v;
                if(dis[v]<dis[u]+e[i].x)
                {
                    dis[v]=dis[u]+e[i].x;
                    if(!b[v]) b[v]=true,q.push(v); 
                }
            }
        }
        return ;
    }
    int main()
    {
        int x,y,z;
        m=read();
        for(int i=1;i<=m;i++)
        {
            x=read(),y=read(),z=read();y++;
            add(x,y,z);
            minn=min(minn,x),maxn=max(maxn,y);
        }
        for(int i=minn;i<=maxn;i++) add(i,i+1,0),add(i+1,i,-1);
        spfa();
        printf("%d",dis[maxn]);
        return 0;
    }
    /*
    我们也可以跑最短路. 
    由约束条件可得
    (1)dis[x]-dis[y+1]<=z.
    (2)0<=dis[i]-dis[i-1]<=1.
    因为是跑最短路.
    所以要把(2)式拆成
    dis[i]-dis[i-1]<=1.
    dis[i-1]-dis[i]<=0.
    然后从终点跑,最后将答案取反. 
    */
    #include<cstring>
    #include<cstdio>
    #include<queue>
    #define MAXN 50001
    using namespace std;
    struct data{int v,next,x;}e[MAXN*3];
    int n,m,head[MAXN],dis[MAXN],cut,maxv,maxn,minn=1e6;
    bool b[MAXN];
    void add(int u,int v,int x)
    {
        e[++cut].v=v;
        e[cut].x=x;
        e[cut].next=head[u];
        head[u]=cut;
    }
    int read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
        return x*f;
    }
    void spfa()
    {
        memset(dis,127/3,sizeof dis);
        queue<int>q;q.push(maxn);dis[maxn]=0;
        while(!q.empty())
        {
            int u=q.front();q.pop();b[u]=false;
            for(int i=head[u];i;i=e[i].next)
            {
                int v=e[i].v;
                if(dis[v]>dis[u]+e[i].x)
                {
                    dis[v]=dis[u]+e[i].x;
                    if(!b[v]) b[v]=true,q.push(v); 
                }
            }
        }
        return ;
    }
    int main()
    {
        int x,y,z;
        m=read();
        for(int i=1;i<=m;i++)
        {
            x=read(),y=read(),z=read();y++;
            add(y,x,-z);
            minn=min(minn,x),maxn=max(maxn,y);
        }
        for(int i=minn;i<=maxn;i++) add(i,i-1,0),add(i,i+1,1);
        spfa();
        printf("%d",-dis[minn]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/nancheng58/p/10068098.html
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