BZOJ 3436 小K的农场 查分约束系统 SPFA判负环

标签：bzoj   查分约束系统   spfa判负环   

题目大意:农场中有一些土地,上面会长一些作物,现在给出一些约束条件,问有没有这种可能.

思路:裸的查分约束系统判负环.记住要跑最长路.

CODE:

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 100010
using namespace std;
 
int points,asks;
int head[MAX],total;
int next[MAX << 1],aim[MAX << 1],length[MAX << 1];
 
int f[MAX],t[MAX];
bool v[MAX];
 
inline void Add(int x,int y,int len)
{
    next[++total] = head[x];
    aim[total] = y;
    length[total] = len;
    head[x] = total;
}
 
bool SPFA()
{
    static queue<int> q;
    memset(f,0xef,sizeof(f));
    f[0] = 0;
    q.push(0);
    while(!q.empty()) {
        int x = q.front(); q.pop();
        v[x] = false;
        for(int i = head[x]; i; i = next[i])
            if(f[aim[i]] < f[x] + length[i]) {
                f[aim[i]] = f[x] + length[i];
                if(!v[aim[i]]) {
                    v[aim[i]] = true;
                    q.push(aim[i]);
                    if(t[aim[i]] = t[x] + 1,t[aim[i]] > points)
                        return false;
                }
            }
    }
    return true;
}
 
int main()
{
    cin >> points >> asks;
    for(int flag,x,y,z,i = 1; i <= asks; ++i) {
        scanf("%d",&flag);
        if(flag == 2) {
            scanf("%d%d%d",&x,&y,&z);
            Add(x,y,-z);
        }
        else if(flag == 1) {
            scanf("%d%d%d",&x,&y,&z);
            Add(y,x,z);
        }
        else {
            scanf("%d%d",&x,&y);
            Add(x,y,0);
            Add(y,x,0);
        }
    }
    for(int i = 1; i <= points; ++i)
        Add(0,i,0);
    if(SPFA())  puts("Yes");
    else    puts("No");
    return 0;
}

BZOJ 3436 小K的农场 查分约束系统 SPFA判负环

标签：bzoj   查分约束系统   spfa判负环   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/41211679