ZOJ 3668 Launching the Spacecraft (差分约束系统，最短路)

题目：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3668

题意：

给一个初始值为0的长度为n的区间，给m个约束l,r,a,b，表示从l到r的区间和>=a且<=b，且每个数的范围为-10000~10000，问这个区间的每个数是多少，要求数尽可能大，且序列字典序最大。

方法：

差分约束系统，以前缀和表示一个节点a[i]

建图：根据下面约束条件建图

a[i]-a[i-1]<=10000

a[i-1]-a[i]<=10000

对于每个条件(l,r,a,b)

a[r]-a[l-1]<=-a

a[l-1]-a[r]<=b

使用最短路尽可能求最大值

(最长路求最小值，最短路求最大值)

void add(int l, int r, int a, int b)
{
    addse(l - 1, r, b);
    addse(r, l - 1, -a);
}
void input()
{
    for (int i = 1; i <= n; i++)
    {
        addse(i, i - 1, 10000);
        addse(i - 1, i, 10000);
    }
    for (int i = 0; i < m; i++)
    {
        int l, r, a, b;
        scanf("%d%d%d%d", &l, &r, &a, &b);
        add(l, r, a, b);
    }
}

代码：

/********************************************
*ACM Solutions
*
*@Title: ZOJ 3668 Launching the Spacecraft
*@Version: 1.0
*@Time: 2014-08-26
*@Solution: http://www.cnblogs.com/xysmlx/p/xxxxxxx.html
*
*@Author: xysmlx(Lingxiao Ma)
*@Blog: http://www.cnblogs.com/xysmlx
*@EMail: xysmlx@163.com
*
*Copyright (C) 2011-2015 xysmlx(Lingxiao Ma)
********************************************/
// #pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <list>
#include <map>
#include <iterator>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long LL;
#define pb push_back
#define ROUND(x) round(x)
#define FLOOR(x) floor(x)
#define CEIL(x) ceil(x)
const int maxn = 1010;
const int maxm = 2000010;
const int inf = 0x3f3f3f3f;
const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double INF = 1e30;
const double eps = 1e-6;
const int P[4] = {0, 0, -1, 1};
const int Q[4] = {1, -1, 0, 0};
const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};
struct Edge
{
    int u, v, w;
    int next;
} edge[maxm];
int head[maxn], en;
void addse(int u, int v, int w)
{
    edge[en].u = u;
    edge[en].v = v;
    edge[en].w = w;
    edge[en].next = head[u];
    head[u] = en++;
}

int n, m;
int d[maxn];
int pre[maxn];//用于解析路径
/*DFS找负环
bool cir[maxn];
void dfs(int u)
{
    cir[u]=true;
    for(int i=head[u]; i!=-1; i=edge[i].next)
        if(!cir[edge[i].v]) dfs(edge[i].v);
}
*/
bool spfa(int s)
{
    int cnt[maxn];
    int mark[maxn];
    queue<int> Q;
    for (int i = 0; i <= n; ++i) d[i] = inf;
    for (int i = 0; i <= n; ++i) pre[i] = i; //用于解析路径
    memset(mark, 0, sizeof(mark));
    memset(cnt, 0, sizeof(cnt));
    d[s] = 0;
    Q.push(s);
    mark[s] = 1;
    cnt[s]++;
    while (Q.size())
    {
        int u = Q.front();
        Q.pop();
        mark[u] = 0;
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            int w = edge[i].w;
            if (d[u] + w < d[v])
            {
                pre[v] = u; // 用于解析路径
                d[v] = d[u] + w;
                if (mark[v] == 0)
                {
                    mark[v] = 1;
                    Q.push(v);
                    if (++cnt[v] > n + 1) return false; //有负环，可以用DFS找
                }
            }
        }
    }
    return true;
}
int kase;
void init()
{
    kase++;
    memset(head, -1, sizeof(head));
    en = 0;
}
void add(int l, int r, int a, int b)
{
    addse(l - 1, r, b);
    addse(r, l - 1, -a);
}
void input()
{
    for (int i = 1; i <= n; i++)
    {
        addse(i, i - 1, 10000);
        addse(i - 1, i, 10000);
    }
    for (int i = 0; i < m; i++)
    {
        int l, r, a, b;
        scanf("%d%d%d%d", &l, &r, &a, &b);
        add(l, r, a, b);
    }
}
void debug()
{
    //
}
void solve()
{
    if (spfa(0))
    {
        for (int i = 1; i <= n; i++)
        {
            if (i < n) printf("%d ", d[i] - d[i - 1]);
            else printf("%d
", d[i] - d[i - 1]);
        }
    }
    else puts("The spacecraft is broken!");
}
void output()
{
    //
}
int main()
{
    // int size = 256 << 20; // 256MB
    // char *p = (char *)malloc(size) + size;
    // __asm__("movl %0, %%esp
" :: "r"(p));

    // std::ios_base::sync_with_stdio(false);
#ifdef xysmlx
    freopen("in.cpp", "r", stdin);
#endif

    kase = 0;
    while (~scanf("%d%d", &n, &m))
    {
        init();
        input();
        solve();
        output();
    }
    return 0;
}