POJ 1201 差分约束系统

题目

Intervals
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 24326 Accepted: 9247
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

题意

给出一些区间，从这些区间里挑一些数。要求每个区间里的数不能小于一定的值。问最少要挑多少个数

题解

定义d[i]为在[0,i]区间里挑了d[i]个数
则有约束条件：
d[i+1]>=d[i],后面的数肯定不会比前面的数少。
d[i]+1>=d[i+1],d[i+1]比d[i]最多多一个1.
d[a[i].R] - d[a[i].L]>=a[i].v,满足题意包含合适的数。
然后就可以用Bllman-ford算法计算结果了。

#define _CRT_SECURE_NO_WARNINGS#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <stack>#include <string>#include <set>#include <cmath>#include <map>#include <queue>#include <sstream>#include <vector>#include <iomanip>#define m0(a) memset(a,0,sizeof(a))#define mm(a) memset(a,0x3f,sizeof(a))#define m_1(a) memset(a,-1,sizeof(a))#define f(i,a,b) for(i = a;i<=b;i++)#define fi(i,a,b) for(i = a;i>=b;i--)#define lowbit(a) ((a)&(-a))#define FFR freopen("data.in","r",stdin)#define FFW freopen("data.out","w",stdout)#define INF 0x3f3f3f3ftypedef long long ll;typedef long double ld;const ld PI = acos(-1.0);using namespace std;#define SIZE ( 50000+10)struct Edge {    int L, R;    int v;};Edge a[SIZE];int d[SIZE];int main() {    //ios_base::sync_with_stdio(false); cin.tie(0);    int n;    while (~scanf("%d", &n)) {        int i;        int Min = INF;        int Max = -INF;        f(i, 1, n) {            int L, R, K;            scanf("%d%d%d", &L, &R, &K);            L++; R++;            Min = min(L - 1, Min);            Max = max(R, Max);            a[i].L = L - 1;            a[i].R = R;            a[i].v = K;        }        m0(d);        int ok = 1;        while (ok) {            ok = 0;            f(i, 1, n) {                if (d[a[i].L] > d[a[i].R] - a[i].v) {                    d[a[i].L] = d[a[i].R] - a[i].v;                    ok = 1;                }            }            f(i, Min, Max - 1)                if (d[i] > d[i + 1]) {                    d[i] = d[i + 1];                    ok = 1;                }            f(i, Min, Max - 1)                if (d[i + 1] > d[i] + 1) {                    d[i + 1] = d[i] + 1;                    ok = 1;                }        }        printf("%d\n", d[Max] - d[Min]);    }    return 0;}