初学差分约束

差分约束系统

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28426		Accepted: 10975

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

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差分约束

引用一个图片，写过spfa应该能秒懂差分约束，SPFA的松弛操作，dist[AC] <= dist[ A->C的其他路 ]即b<=a+c

另一个问题有一系列不等式

B - A <= c

C - B <= a

C - A <= B

把减法移到另一边，就和我们的松弛操作很像了

详情

http://blog.csdn.net/consciousman/article/details/53812818

题意：给定n个区间，[ai,bi]这个区间至少选选出ci个整数，求一个最小集合z，满足每个区间的要求，输出集合z的大小。

分析：令d[i]表示0到i这个区间内至少要选出d[i]个数，那么对于每个[ai,bi],有d[b] - d[ai-1] >= ci，同时隐含的一个条件是0 <= d[i] - d[i-1] <= 1,但是因此d[-1]不能表示，令d[i+1]示0到i这个区间内至少要选出d[i+1]个数，然后d[0] = 0,直接求取最长路就行了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>

using namespace std;
const int maxn = 5e4 + 7;
const int maxm = maxn << 2;
const int inf = ~0U>>1;

int n, a, b, c, first[maxn], sign;

struct Node {
    int to, w, next;
} edge[maxm];

void init()
{
    ///注意这里的maxn
    for(int i = 0; i < maxn; i ++ ) {
        first[i] = -1;
    }
    sign = 0;
}

void add_edge(int u, int v, int w)
{
    edge[sign].to = v;
    edge[sign].w  = w;
    edge[sign].next = first[u];
    first[u] = sign ++;
}

int dist[maxn], inq[maxn], L, R;

void SPFA()
{
    for(int i = L; i <= R; i ++ ) {
        dist[i] = -inf, inq[i] = 0;
    }
    queue<int> Q;
    Q.push(L);
    inq[L] = 1, dist[L] = 0;
    while(!Q.empty()) {
        int now = Q.front();
        Q.pop();
        inq[now] = 0;
        for(int i = first[now]; ~i; i = edge[i].next) {
            int to = edge[i].to, w = edge[i].w;
            if(dist[to] < dist[now] + w) {
                dist[to] = dist[now] + w;
                if(!inq[to]) {
                    Q.push(to);
                    inq[to] = 1;
                }
            }
        }
    }
}

int main()
{
    while(~scanf("%d", &n)) {
        init();
        L = inf, R = -inf;
        /**
       约束不等式
       d[ bi ] - d[ ai - 1] >= ci;
       d[i] - d[i - 1] >= 0;
       d[i] - d[i - 1] >= -1;
        */
        for(int i = 1; i <= n; i ++ ) {
            scanf("%d %d %d", &a, &b, &c);
            a ++, b ++;
            add_edge(a - 1, b, c);
            L = min(L, a - 1);
            R = max(R, b);
        }
        for(int i = L; i < R; i ++ ) {
            add_edge(i, i + 1, 0);
            add_edge(i + 1, i, -1);
        }
        SPFA();
        printf("%d
", dist[R]);
    }
    return 0;
}