POJ 1201

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第一次接触差分约束系统，现在只是知道了一个差分约束的公式就是最短路的最终形式 v-u<=w，在最短路求解中只要是满足条件 if（dis[u]+w<dis[v]）就将dis[v]的值更新，最后都会满足约束条件 v-u<=w。

题目说[ai, bi]区间内和点集Z至少有ci个共同元素，那也就是说如果我用Si表示区间[0,i]区间内至少有多少个元素的话，那么Sbi - Sai >= ci,这样我们就构造出来了一系列边，权值为ci，但是这远远不够，因为有很多点依然没有相连接起来（也就是从起点可能根本就还没有到终点的路线）,此时，我们再看看Si的定义，也不难写出0<=Si - Si-1<=1的限制条件，虽然看上去是没有什么意义的条件，但是如果你也把它构造出一系列的边的话，这样从起点到终点的最短路也就顺理成章的出现了。

我们将上面的限制条件写为同意的形式：
Sai-1 - Sbi <= -ci
Si - Si-1 >= 0
Si-1 - Si >= -1

Intervals

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 26017 Accepted: 9943

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

Source

Southwestern Europe 2002

#include<stdio.h>#include<string.h>#include<queue>#define MAX 512345#define INF 0x3f3f3f3fusing namespace std;struct Node{    int v,w,next;}arr[MAX];int head[MAX];bool vis[MAX];int dis[MAX];int minx=INF;int maxx=-INF;int n;int cnt;void Init(){    memset(head,-1,sizeof(head));    memset(vis,true,sizeof(vis));    memset(dis,INF,sizeof(dis));    cnt=0;}void add(int u,int v,int w){    arr[cnt].v=v;    arr[cnt].w=w;    arr[cnt].next=head[u];    head[u]=cnt++;}void spfa(){    dis[maxx]=0;    vis[maxx]=false;    queue<int>que;    que.push(maxx);    int temp;    while(!que.empty())    {        temp=que.front();        que.pop();        vis[temp]=true;        for(int i=head[temp];i!=-1;i=arr[i].next)        {            int xxx=dis[temp]+arr[i].w;            if(dis[arr[i].v]>xxx)            {                dis[arr[i].v]=xxx;                if(vis[arr[i].v])                {                    vis[arr[i].v]=false;                    que.push(arr[i].v);                }            }        }    }}int main(){    Init();    scanf("%d",&n);    int u,v,w;    for(int i=0;i<n;i++)    {        scanf("%d%d%d",&u,&v,&w);        add(v+1,u,-w);        if(maxx<v+1)            maxx=v+1;        if(minx>u)            minx=u;    }    for(int i=minx;i<maxx;i++)    {        add(i+1,i,0);        add(i,i+1,1);    }    spfa();    printf("%d\n",-dis[minx]);    return 0;}

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