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 0x7fffffffusing 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;}