差分约束

例题1 interval

  • 最长路
  • 数组下标为-1怎么处理
  • ”标准输出太巨怎么办“
/*
reference:
	
translation:
	
solution:
	
1---->在a[i]和b[i]+1之间连一条长度为c[i]的有向边

2---->在i和i+1之间连一条长度为0的有向边

3---->在i+1和i之间连一条长度为-1的有向边
trigger:
	
note:
	*我们在i-1和i之间连长度为0的有向边,在i和i-1之间连长度为-1的有向边
因为i的范围是0-50000所以i-1会变成负数,我们都知道,若数组的下标为负,程序会运行
错误,所以我们不妨给所有的变量加1,这样就不会出现负下标了
date:
	2019.09.05
*/
const int N=1e6+10;

int head[N],edge_num,maxx=INT_MIN,minn=INT_MAX;
int n,m,s;
int dis[N];
bool vis[N];

struct edge{
	int v,w,next;
}e[N];

inline void add(int u,int v,int w){e[++edge_num].v=v;e[edge_num].w=w;e[edge_num].next=head[u];head[u]=edge_num;}

queue<int>q;
inline void spfa(){
	mem(dis,-0x3f);
	mem(vis,0);
	q.push(0);
	vis[0]=1;
	dis[0]=0;
	while(!q.empty()){
		int u=q.front();q.pop();
		vis[u]=0;
		ee(i,u){
			int v=e[i].v,w=e[i].w;
			if(dis[v]<dis[u]+w){
				dis[v]=dis[u]+w;
				if(!vis[v]){
					q.push(v);
					vis[v]=1;
				}
			}
		}
	}
} 

#undef int
int main(){
#define int long long
	#ifdef WIN32
	freopen("interval.txt","r",stdin);
	#endif
	int T;rd(T);
	rep(cases,1,T){
		//初始化;
		mem(e,0);mem(head,0);
		edge_num=0,maxx=INT_MIN,minn=INT_MAX;
		rd(m);
		while(m--){
			int b,e,t;rd(b),rd(e),rd(t);
			add(b,e+1,t);
			maxx=max(maxx,e+1);
		} 
		rep(i,0,maxx){
			add(i+1,i,-1);
			add(i,i+1,0);
		}
		spfa();
		if(cases<T)printf("%lld
",dis[maxx]);
		else printf("%lld",dis[maxx]);
	}
    return 0;
}

例二(看似很裸)poj 3159 candies

  • poj 交的时候不能在文本末尾写注释
  • spfa+queue竟然超时是什么鬼!!!,spfa只能加数组才能过!!!连加双端队列优化都不可以………………仙人板板
/*
reference:
	
translation:
	
solution:

trigger:
	
note:
	*
date:
	2019.09.04
*/
#include<iostream>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define dwn(i,a,b) for(int i=a;i>=b;--i) 
template <typename T> inline void rd(T &x){x=0;char c=getchar();int f=0;while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}x=f?-x:x;} 
#define ee(i,x) for(int i=head[x];i;i=e[i].next)
#define mem(a,b) memset(a,b,sizeof(a))

const int N = 30010,M = 150010;

int head[N],edge_num;
int n,m,dis[N];
bool vis[N];

struct edge{
	int v,w,next;
}e[M];

inline void add(int u,int v,int w){e[++edge_num].v=v;e[edge_num].w=w;e[edge_num].next=head[u];head[u]=edge_num;}
//inline void adde(int u,int v){e[++edge_num].v=v;e[edge_num].next=head[u];head[u]=edge_num;}
/*
deque<int>q;
inline void spfa(){
	mem(dis,0x3f);
	mem(vis,0);
	q.push_back(1);
	vis[1]=1;
	dis[1]=0;
	while(!q.empty()){
		int u=q.front();q.pop_front();
		vis[u]=0;
		ee(i,u){
			int v=e[i].v,w=e[i].w;
			if(dis[v]>dis[u]+w){
				dis[v]=dis[u]+w;
				if(q.empty() || dis[v]<dis[u])
					q.push_front(v);
				else q.push_back(v);
				vis[v]=1;
			}
		}
	}
}*/

int q[N];
void spfa(int start,int n)
{
    int top=0;
    for(int v=1;v<=n;v++)//初始化
    {
        if(v==start)
        {
            q[top++]=v;
            vis[v]=true;
            dis[v]=0;
        }
        else
        {
            vis[v]=false;
            dis[v]=0x3f3f3f3f;
        }
    }
    while(top!=0){
        int u=q[--top];
        vis[u]=0;
        ee(i,u){
            int v=e[i].v,w=e[i].w;
            if(dis[v]>dis[u]+w){
                dis[v]=dis[u]+w;
                if(!vis[v]){
                    vis[v]=true;
                    q[top++]=v;
                }
            }
        }
    }
}

int main(){
	rd(n),rd(m);
	while(m--){
		int u,v,w;rd(u),rd(v),rd(w);
		add(u,v,w);
	}
	spfa(1,n);
	printf("%d
",dis[n]);
    return 0;
}
/*
2 2
1 2 5
2 1 4
*/
//5
原文地址:https://www.cnblogs.com/sjsjsj-minus-Si/p/11634692.html