LIS ZOJ - 4028

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4028

memset超时

这题竟然是一个差分约束

好吧呢

对于每一个a[i], l <= a[i] <= r

那么设一个源点s

使 l <= a[i] - s <= r  是不是就能建边了

然后对于每一个f[i]

如果前面有一个相等的f[j]

则肯定 a[i] <= a[j]  又能建边了

根据LIS的传递关系

对于每个f[i] 肯定是由上一个等级的传递过来的

即 a[i] > a[j]   是不是又能建边了

不会建边？

请移步： https://www.cnblogs.com/WTSRUVF/p/9153758.html

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 110000, INF = 0x7fffffff;
int head[maxn], vis[maxn];
LL d[maxn];
int cnt;
int n, m, s;
int ans[maxn];

struct node
{
    int u, v, next;
    int w;
}Node[1000500];

void add(int u, int v, int w)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].w = w;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void init()
{
    for(int i = 0; i <= n + 1; i++)
    {
        head[i] = -1;
        ans[i] = 0;
        vis[i] = 0;
        d[i] = INF;
    }
    cnt = 0;
}

bool spfa()
{
    deque<int> Q;
    Q.push_front(s);
    d[s] = 0;
    vis[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop_front();
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = Node[i].next)
        {
            int v = Node[i].v;
            if(d[v] > d[u] + Node[i].w)
            {
                d[v] = d[u] + Node[i].w;
                if(!vis[v])
                {
                    if(Q.empty()) Q.push_front(v);
                    else if(d[v] < d[Q.front()]) Q.push_front(v);
                    else Q.push_back(v);
                    vis[v] = 1;
                    if(++ans[v] > n) return 1;
                }
            }
        }
    }
    return 0;
}

int pre[maxn];

int main()
{
    int T;
    rd(T);
    while(T--)
    {

        rd(n);
        init();
        s = n + 1;
        mem(pre, 0);
        for(int i = 1; i <= n; i++)
        {
            int f;
            rd(f);
            if(pre[f]) add(pre[f], i, 0);
            if(f > 0) add(i, pre[f - 1], -1);
            pre[f] = i;
        }
        for(int i = 1; i <= n; i++)
        {
            int l, r;
            rd(l), rd(r);
            add(s, i, r);
            add(i, s, -l);
        }
        spfa();
        for(int i = 1; i <= n; i++)
        {
            if(i != 1) printf(" ");
            printf("%lld", d[i]);

        printf("\n");
        }

    }

    return 0;
}

原文地址：https://www.cnblogs.com/WTSRUVF/p/10776950.html