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
/*又回過頭來看了一下半年前學的差分約束.感覺自己還是弱弱的. 由約束條件可得(1)dis[y+1]-dis[x]>=z.(2)0<=dis[i]-dis[i-1]<=1.因為是跑最長路.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;}/*我們也可以跑最短路. 由約束條件可得(1)dis[x]-dis[y+1]<=z.(2)0<=dis[i]-dis[i-1]<=1.因為是跑最短路.所以要把(2)式拆成dis[i]-dis[i-1]<=1.dis[i-1]-dis[i]<=0.然后從終點跑,最后將答案取反. */#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;}新聞熱點
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