Luogu1993 小K的农场 (差分约束)

时间:2019-08-24
本文章向大家介绍Luogu1993 小K的农场 (差分约束),主要包括Luogu1993 小K的农场 (差分约束)使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

\(if \ a - b <= c, AddEdge(b, a, c)\)
Be careful, MLE is not good.

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a))
#define nR(a,b,c) for(register int a = (b); (a) >= (c); --(a))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))

#define ON_DEBUGG

#ifdef ON_DEBUGG

#define D_e_Line printf("\n----------\n") 
#define D_e(x) cout << (#x) << " : " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt", "r", stdin)

#else

#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;

#endif
using namespace std;
struct ios{
    template<typename ATP>inline ios& operator >> (ATP &x){
        x = 0; int f = 1; char ch;
        for(ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
        while(ch >= '0' && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();
        x *= f;
        return *this;
    }
}io;

const int N = 20007;

struct Edge{
    int nxt, pre, w;
}e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v, int w){
    e[++cntEdge] = (Edge){ head[u], v, w}, head[u] = cntEdge;
}

int vis[N], dis[N];
inline bool SPFA(int u){
    vis[u] = true;
    for(register int i = head[u]; i; i = e[i].nxt){
        if(dis[e[i].pre] > dis[u] + e[i].w){
            dis[e[i].pre] = dis[u] + e[i].w;
            if(vis[e[i].pre] || SPFA(e[i].pre)){
                return true;
            }
        }
    }
    vis[u] = false;
    return false;
}

int main(){
//FileOpen();
    int n, m;
    io >> n >> m;
    
    R(i,1,m){
        int opt;
        io >> opt;
        if(opt == 1){
            int x, y, w;
            io >> x >> y >> w;
            add(x, y, -w);
        }
        else if(opt == 2){
            int x, y, w;
            io >> x >> y >> w;
            add(y, x, w);   
        }
        else if(opt == 3){
            int x, y;
            io >> x >> y;
            add(x, y, 0);
            add(y, x, 0);
        }
    }
    
    R(i,1,n){
        add(0, i, 0); // this sentence caused MLE !
        dis[i] = 0x3f3f3f3f;
    }
    
    if(SPFA(0) == false){
        printf("Yes\n");
    }
    else{
        printf("No\n");
    }
    return 0;
}

原文地址:https://www.cnblogs.com/bingoyes/p/11406299.html