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 50001using 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 50001using 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;}