POJ1201--差分约束--Intervals

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

53 7 38 10 36 8 11 3 110 11 1

Sample Output

6

/*差分约束问题。不知道是什么的可以先看我的博客，转载的。。此题有50000条边。一：d[t]-d[s-1]>=key  ==>  d[s-1]-d[t]<=-key二：d[i]-d[i-1]<=1;三：d[i-1]-d[i]<=0;差分约束问题用最短路来解决我用的是bellmand-ford算法枚举每条边，if(d[edge[i].s]-d[edge[i].t]<edge[i].w){d[edge[i].t]=d[edge[i].s]+edge[i].w;}现在我们着重想一想初始化的问题，刚开始的时候我们一定要让d[edge[i].s]-d[edge[i].t]<edge[i].w;==>我是说第一步的时候。这样才会走出去。类似于我们做裸的最短路题的时候，起点的距离置0，其他置无穷大。哦，对了。这题我们读入边的时候同时把最大的点和最小的点存下来因为等下就是求这两点的最短路额。。说回刚才的问题，要第一步可走，edge[i].w都非正。我们让d[mi-1]=inf;则如果d[edge[i].s]-d[edge[i].t]>edge[i].w就让d[edge[i].s]=d[edge[i].t]+edge[i].w;显然第一步能走出去吧。。*/#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define maxn 50008#define inf 0x3f3f3f3fint d[maxn],n,mi,ma;inline int max(int a,int b){return a>b?a:b;}inline int min(int a,int b){return a>b?b:a;}struct Edge{int s,t,w;}edge[maxn];void bellmand(){bool flag=true;while(flag){flag=false;for(int i=1;i<=n;i++){if(d[edge[i].s]-d[edge[i].t]>edge[i].w){d[edge[i].s]=d[edge[i].t]+edge[i].w;flag=true;}}for(int i=mi;i<=ma;i++){if(d[i]>d[i-1]+1){d[i]=d[i-1]+1;flag=true;}}for(int i=ma;i>=mi;i--){if(d[i-1]>d[i]){d[i-1]=d[i];flag=true;}}}}int main(){while(scanf("%d",&n)!=EOF){int u,v,w;mi=inf,ma=0;for(int i=1;i<=n;i++){scanf("%d%d%d",&u,&v,&w);edge[i].s=u;edge[i].t=v;edge[i].w=-w;edge[i].s--;mi=min(mi,u);ma=max(ma,v);}memset(d,0,sizeof(d));d[mi-1]=inf;bellmand();printf("%d\n",d[ma]-d[mi-1]);}return 0;}