差分约束（刷题）

差分约束#

来自Kersen

给定 $n$ 个变量和 $m$ 个不等式，每个不等式形如 $x[i] - x[j] <= a[k]$，求 $x[n-1] - x[0]$ 的最大值。
$(0 <= i, j < n)$
例：
$n = 4$
$x_1 − x_0 <= 2$ ①
$x_2−x_0<=7$ ②
$x_3−x_0<=8$ ③
$x_2−x_1<=3$ ④
$x_3−x_2<=2$ ⑤
然后经过认真的瞎搞计算就变成了这个鸭子：
1、③ $x_3−x_0 <= 8$
2、②+⑤ $x_3−x_0 <= 9$
3、①+④+⑤ $x_3−x_0 <= 7$

对于每个不等式 $x_i−x_j<=a_k$，对结点 $j$ 和 $i$ 建立一条 $j−>i$ 的有向边，
边权为 $a_k$，求 $x_{n−1}−x_0$ 的最大值就是求 $0$ 到 $n−1$ 的最短路

求最大值的时候把全部的不等式都化成：$x[i] - x[j] <= w$ 的形式;

然后建由 $j$ 指向 $i$ 边权为 $w$ 的边，跑最短路；

求最小值的时候把全部的不等式都化成：$$

差分约束系统的解有两种特殊情况：#

1.$x[t] - x[s]$ 最大值不存在

存在负环，用 $spfa$ 判负环

2.$x[t] - x[s]$ 的值无限大

图不连通…………

不等式标准化#

1.如果给出的既有 $>=$ 又有 $<=$

如果需要求的是两个变量差的最大值，那么需要将所有不等式转变成"<="的形式，建图后求最短路；
相反，如果需要求的是两个变量差的最小值，那么需要将所有不等式转化成">="，建图后求最长路。

2.如果有形如 $A - B = C$ 的等式

我们可以将它转化成以下两个不等式：
$A − B >= c$ ①
$A − B <= c$ ②
再通过上面的方法将其中一种不等号反向，建图即可。

3.如果遇到不带等号的，$a + b > c$

我们可以将它转化成"$<=$"或者"$>=$"的形式，即 $A−B<=c−1$ 形式就可以做了...

Intervals#

给定多个区间，要求每个区间选一定数量的数，求所选的数至少为多少

solution#

差分约束板子

可以推出下面的不等式

$a-b >= c$

$x[i] - [i-1] >= 0$

$S_i-S_{(i-1)}>=0$

$S_{(i-1)}-S{i}>=-1$

/*
work by:Ariel_
*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <iostream>
#include <cstring>
using namespace std;
const int M = 1e6 + 11;
int read(){
   int x = 0,f = 1;char c = getchar();
   while(c < '0'||c > '9'){if(c == '-')f = -1;c = getchar();}
   while(c >= '0'&&c <= '9'){x = x*10 + c - '0';c = getchar();}
   return f*x;
}
struct edge{
  int v,nxt,w;
}e[M << 1];
int cnt,head[M];
void add_edge(int u,int v,int w){
	e[++cnt] = (edge){v, head[u], w};
	head[u] = cnt;
}
int T, n, maxn = -0x3f3f3f3f,dis[M],vis[M];
queue<int> q;
void spfa(){
    memset(dis, -0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    dis[0] = 0,vis[0] = 1;
    q.push(0);
    while(!q.empty()){
       int u = q.front();
       q.pop();vis[u] = 0;
       for(int i = head[u]; i;i = e[i].nxt){
       	   int v = e[i].v;
       	   if(dis[v] < dis[u] + e[i].w){
       	   	   dis[v] = dis[u] + e[i].w;
       	   	   if(!vis[v]){
       	   	   	  vis[v] = 1;
       	   	   	  q.push(v); 
				}
			}
	   }
	}
}
int main(){
   n = read();
   	 for(int i = 1,a,b,c;i <= n; i++){
   	   a = read(),b = read(),c = read();
       add_edge(a, b + 1, c);
	    maxn = max(b + 1,maxn);
     }
    for(int i = 1;i <= maxn; i++){
   	   add_edge(i - 1, i, 0);
   	   add_edge(i,i - 1, -1);
    }
     spfa();
     printf("%d\n", dis[maxn]);
  
}

出纳员问题#

题意较长……没时间写了，看链接把

solution#

/*
work by:Ariel_
*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int M = 1e5 + 10;
const int eqs = 1e-10;
int read(){
   int x = 0,f = 1;char c = getchar();
   while(c < '0'||c > '9'){if(c == '-')f = -1;c = getchar();}
   while(c >= '0'&&c <= '9'){x = x*10 + c - '0';c = getchar();}
   return f*x;
}
int T,r[25],a[1010],n,dis[100010],num[100010];
bool vis[100010];
struct edge{
  int v, nxt, w;	
}e[M];
int js,head[M];
void add_edge(int u,int v,int w){
	e[++js] = (edge){v,head[u],w};
	head[u] = js;
}
void clear(){
	js = 0;
	memset(head, 0, sizeof(head));
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
} 
queue<int>q;
bool check(int x){
	clear();
	add_edge(0, 24, x);
    for (int i = 1; i <= 24; i++){
      add_edge(i, i - 1, 0);
      add_edge(i - 1, i, num[i]);
      if(i > 8) add_edge(i, i - 8, -r[i]);
      else add_edge(i, i + 16, x - r[i]);
    }
	q.push(0), dis[0] = 0, vis[0] = 0;
    while(!q.empty()){
      int u = q.front();
      q.pop();
      vis[u] = 0;
      for (int i = head[u]; i; i = e[i].nxt)
      {
        int v = e[i].v, w = e[i].w;
        if(dis[v] > dis[u] + w){
          dis[v] = dis[u] + w;
          js++;
          if(!vis[v]){
            vis[v] = 1;
            q.push(v);
          }
          if(js > 2000) return 0;
        }
      }
    }
    return dis[24] == x;
}
int main(){
   T = read();
   while(T--){memset(num, 0, sizeof(num));
   	 for (int i = 1;i <= 24; i++)r[i] = read();
   	   n = read();
   	 for (int i = 1, x;i <= n; i++){
   	 	 x = read() + 1;
   	 	 num[x]++;
	  }
    int l = 0, r = n, ans = 0x3f3f3f3f;
     while(l <= r){
       int mid = (r + l) >> 1;
       if(check(mid)) ans = mid, r = mid - 1;
      else l = mid + 1;
     }
    if(ans == 0x3f3f3f3f) cout << "No Solution" << endl;
    else cout << ans << endl;
   }
}

糖果#

一大串不等式关系，跑差分约束

/*
work by:Ariel_
*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define int long long
const int M = 1e6 + 10;
int read(){
   int x = 0,f = 1;char c = getchar();
   while(c < '0'||c > '9'){if(c == '-')f = -1;c = getchar();}
   while(c >= '0'&&c <= '9'){x = x*10 + c - '0';c = getchar();}
   return f*x;
}
struct edge{
   int v,nxt,w;
}e[M << 1];
int cnt,head[M];
void add_edge(int u,int v,int w){
	e[++cnt] = (edge){v, head[u], w};
	head[u] = cnt;
}
int n, k, dis[M], ans, tot[M];
bool vis[M];
signed main(){
  n = read(),k = read();
   for(int i = 1, x, a, b;i <= k; i++){
   	  x = read(),a = read(),b = read();
   	  if(x == 1)add_edge(a, b, 0),add_edge(b, a, 0);
	  if(x == 2){
	  	if(a == b){
	  		printf("-1");return 0 ;
		  }
		add_edge(a, b, 1);
	  }
	  if(x == 3)add_edge(b, a, 0);
	  if(x == 4){
	    if(a == b){
	  		printf("-1");return 0;
		  }
	  	add_edge(b, a, 1);
	  }
	  if(x == 5)add_edge(a, b, 0); 
   }
  for(int i = n; i >= 1; i--)add_edge(0, i, 1);
   std::queue<int> q;
    q.push(0);
	vis[0] = 1,dis[0] = 0;
	 while(!q.empty()){
	   int u = q.front();q.pop();
	   vis[u] = 0;
	   if(tot[u]== n - 1){
	      printf("-1");return 0;
	   }
	   tot[u]++;
	   for(int i = head[u]; i;i = e[i].nxt){
   	   	  int v = e[i].v;
   	   	 if(dis[v] < dis[u] + e[i].w){
   	   	    dis[v] = dis[u] + e[i].w;
			if(!vis[v])vis[v] = 1,q.push(v);  		
		  }
	   }
	}
  int ans = 0;
  for(int i = 1;i <= n; i++) ans += dis[i];
   printf("%lld", ans);
}

布局 Layout#

既有 $>=$ 又有 $<=$ 处理如上

两种特殊情况

/*
work by:Ariel_
*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int M = 10100;
int read(){
   int x = 0,f = 1;char c = getchar();
   while(c < '0'||c > '9'){if(c == '-')f = -1;c = getchar();}
   while(c >= '0'&&c <= '9'){x = x*10 + c - '0';c = getchar();}
   return f*x;
}
struct edge{ 
  int v, nxt, w;
}e[M << 1];
int js,head[M];
void add_edge(int u,int v,int w){
	e[++js] = (edge){v,head[u],w};
	head[u] = js;
}
int dis[M],tot[M], n, ml,md;
bool vis[M];
int spfa(int x){
   memset(dis,0x7f/3,sizeof(dis));
   memset(vis,0,sizeof(vis));
   memset(tot,0,sizeof(tot));
   queue<int> q;
   vis[x] = 1;q.push(x);dis[x] = 0;
   while(!q.empty()){
   	  int u = q.front();q.pop();
   	  vis[u] = 0;tot[u]++;
   	  if(tot[u] > n) return -1;
   	  for(int i = head[u]; i; i = e[i].nxt){
		    int v = e[i].v;
		    if(dis[v] > dis[u] + e[i].w){
		    	dis[v] = dis[u]+ e[i].w;
		    	if(!vis[v]){
		    	   q.push(v);
				   vis[v] = 1; 
				}
			}
		}
   }
   if(dis[n] > 1e8)return -2;
   return dis[n];
}

int main(){
   n = read(),ml = read(),md = read();
   for(int i = 1,u, v, w;i <= ml; i++){
   	  u = read(),v = read(),w = read();
   	  add_edge(u, v, w); 
   }
   for(int i = 1, u, v, w;i <= md; i++){
   	  u = read(),v = read(),w = read();
   	  add_edge(v, u, -w);
   }
   for(int i = 1;i < n; i++)add_edge(i + 1, i, 0);
   for(int i = 1;i <= n; i++)add_edge(0, i, 0);
   if(spfa(0) <= -1)printf("%d",spfa(spfa(0)));
   else printf("%d",spfa(1));
}