HDUoj 1384 Intervals 差分约束

Intervals

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem 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 endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

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

题意：给n个[st,en]区间，每个区间取出val个数，问至少能取出多少数。

差分约束，设total[i]为区间[0,i]最多要取多少数。所以根据区间可以写出
total[en]-total[st-1] >= val，题中包含一个隐含的不等式:
0<=total[i]-total[i-1]<=1。根据这三个不等式就可以建图并切跑最长路并输出dis[minen];
跑最长路的原因是：
①：对于差分不等式，a - b <= c ，建一条 b 到 a 的权值为 c 的边，求的是最短路，得到的是最大值
②：对于不等式 a - b >= c ，建一条 b 到 a 的权值为 c 的边，求的是最长路，得到的是最小值。
这里找的是最小值，所以跑最长路。

CODE

#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"string.h"
#include"math.h"
#include"stdlib.h"
#include"queue"
#define maxn 200010
#define inf 0x7fffffff
using namespace std;

struct node
{
    int en;
    int val;
    int next;
}edge[maxn];  ///边
int n,len,minst,minen; ///len 边的数量，minst 起点，minen 终点
int dis[maxn];  ///spfa用
int head[maxn];  ///头结点
int vis[maxn];  ///spfa用

void init()  ///初始化
{
    minst = inf;
    minen = -1;
    len = 0;
    for(int i = 0;i < maxn;i++)
    {
        dis[i] = -inf;
        head[i] = -1;
        vis[i] = false;
    }
}

void add(int st,int en,int val) ///邻接表加边
{
    edge[len].en = en;
    edge[len].val = val;
    edge[len].next = head[st];
    head[st] = len++;
}

void spfa()  ///spfa模版
{
    dis[minst] = 0; 
    queue<int>q;
    q.push(minst);   ///起点进队
    vis[minst] = true;
    while(!q.empty())
    {
        int t = q.front();
        q.pop();
        vis[t] = false;
        ///printf("a");
        for(int i = head[t];i != -1;i = edge[i].next)
        {
            node e = edge[i];
            if(dis[e.en] < dis[t]+e.val)
            {
                dis[e.en] = dis[t]+e.val;
                if(!vis[e.en])
                {
                    vis[e.en] = true;
                    q.push(e.en);
                }
            }
        }
    }
}

int main(void)
{
    while(scanf("%d",&n) !=EOF)
    {
        init();
        for(int i = 1;i <= n;i++)   ///total[en]-total[st-1] >= val
        {
            int st,en,val;
            scanf("%d%d%d",&st,&en,&val);
            add(st-1,en,val);
            if(minst > st-1) minst = st-1;
            if(minen < en) minen = en;
        }
        for(int i = minst;i <= minen;i++)
        {
            add(i-1,i,0);  ///total[i]-total[i-1] >= 0
            add(i,i-1,-1);  ///total[i-1]-total[i] >= -1
        }
        spfa();
        /*for(int i = minst;i <= minen;i++) printf("%d ",dis[i]);*/
        printf("%d\n",dis[minen]);
    }
    return 0;
}