差分约束系统

解题步骤：

1.找出题目约束条件

2.把约束条件转化为 A - B <= C 的形式

3.连 B->A 权值为C

4.跑一遍最短路

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模版：

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

#define N 100000
#define M N<<2
#define Inf 0x3f3f3f3f

int nxt[M], to[M], vl[M], head[N];
int dist[N], in[N];
bool vis[N];
int n, cnt = 0;

void addeage( int u, int v, int w ){
    nxt[ ++ cnt ] = head[u], to[cnt] = v, vl[cnt] = w, head[u] = cnt;
}

int SPFA(){
    queue<int> q;
    for ( int i = 1; i <= n; i++) dist[i] = Inf;
    memset( vis, false, sizeof(vis));
    memset( in, 0, sizeof(in));
    dist[1] = 0;
    vis[1] = true;
    in[1]++;
    while ( !q.empty() ) q.pop();
    q.push(1);
    while ( !q.empty() ){
        int cur = q.front();
        q.pop();
        vis[ cur ] = false;
        for ( int i = head[cur]; i != -1; i = nxt[i] ){
            int v = to[i];
            if ( dist[v] > dist[cur] + vl[i] ){
                dist[v] = dist[cur] + vl[i];
                if ( !vis[v] ){
                    q.push(v);
                    if ( ++in[v] > n ) return -1;
                    vis[v] = true;
                }
            }
        }
    }
    if ( dist[n] == Inf ) return -2;
    return dist[n];
}


int main(){
    addeage();
    SPFA();
    return 0;
}