zoj4028 LIS

差分约束瞎搞一下，话说这个数据不知道怎么回事，我的图按道理而言最多只有4n条边，开5n还不够？？必须6n？？

约束条件首先根据f函数可建立两点之间的约束，不妨设d[i]为i到0的距离，则对于f[i] == f[j] + 1(i>j)的情况，必定d[i] - d[j] >= 1,对于f[i] == f[j]的情况，必定d[j] - d[i] >= 0(i>j)

区间范围就不说了吧。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<string>
#include<set>
#include<algorithm>
#include<vector>
#include<queue>
#include<list>
#include<cmath>
#include<cstring>
#include<map>
#include<stack>
using namespace std;
#define check(x,y) (x>=5&&y>=5&&x<n+5&&y<m+5)
#define dight(chr) (chr>='0'&&chr<='9')
#define alpha(chr) (chr>='a'&&chr<='z')
#define rep(i,a,n) for(int i=a;i<n;++i)
#define repe(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define loop(a) for(int i=0;i<a;++i)
#define loope(a) for(int i=1;i<=a;++i)
#define clc(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f3f3f3f
#define maxn 100005
#define ull unsigned long long
#define ll long long
#define hashmod 99999839
#define mod 1000000000
#define repe(x,y,i) for(i=(x);i<=(y);++i)
#define repne(x,y,i) for(i=(x);i<(y);++i)
#define MAX(x,y) (x) < (y) ? (y) : (x);
struct edge{
    int to;
    int cost;
    int next;
}G[600005];
int f,l,r,p[maxn],len;
int head[maxn],n;
ll dis[maxn];
bool inq[maxn];
void Read(int &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-1;
    for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;}
void Read(ll &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if
    (chr=='-')sign=-1;
    for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;}
void spfa(int st){
    queue<int> q;
    memset(inq,false,sizeof(inq));
    for(int i = 0;i <= n;++i) dis[i] = INF;
    q.push(st);
    inq[st] = true;
    dis[st] = 0;
    while(!q.empty()){
        int v = q.front();
        q.pop();
        inq[v] = false;
        for(int i = head[v];i;i = G[i].next){
            edge& t = G[i];
            if(dis[t.to] > dis[v] + t.cost){
                dis[t.to] = dis[v] + t.cost;
                if(!inq[t.to]) q.push(t.to);
            }
        }
    }
    for(int i = 1;i < n;++i) printf("%lld ",dis[i]);
    printf("%lld
",dis[n]);
}
int main(){
  ////  freopen("a.in","r",stdin);
  //  freopen("b.out","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        len = 1;
        memset(head,0,sizeof(head));
        memset(p,0,sizeof(p));
        Read(f);
        p[f] = 1;
        for(int i = 2;i <= n;++i){
            Read(f);
            if(f == 1){
                G[len].to = i,G[len].cost = 0,G[len].next = head[p[1]],head[p[1]] = len,++len;
                p[f] = i;
                continue;
            }
            int t = p[f - 1];
            G[len].to = t,G[len].cost = -1,G[len].next = head[i],head[i] = len,++len;
            t = p[f];
            if(t) G[len].to = i,G[len].cost = 0,G[len].next = head[t],head[t] = len,++len;
            p[f] = i;
        }
        for(int i = 1;i <= n;++i){
            Read(l),Read(r);
            G[len].to = 0,G[len].cost = -l,G[len].next = head[i],head[i] = len,++len;
            G[len].to = i,G[len].cost = r,G[len].next = head[0],head[0] = len,++len;
        }
        spfa(0);
    }
    return 0;
}