BZOJ 3436: 小K的农场 差分约束

题目链接：

http://www.lydsy.com/JudgeOnline/problem.php?id=3436

题解：

裸的差分约束：

1、a>=b+c  ->  b<=a-c  ->  d[v]<=d[u]+w  ->  建一条边从a到b，权值为-c

2、a<=b+c  ->  d[v]<=d[u]+w  -> 建一条边从b到a，权值为c

3、a==b  ->  d[v]<=d[u]+0&&d[u]<=d[v]+0 建一条边从a到b，权值为0。建一条边从b到a,权值为0.

建完图，从原点0到各点连边，权值都为0，跑最短路。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<queue>
#define mp make_pair
#define X first
#define Y second
using namespace std;

const int maxn = 10000 + 10;

vector<pair<int,int> > G[maxn];

int n, m;

int inq[maxn], d[maxn], cnt[maxn];
bool spfa(int s) {
    memset(inq, 0, sizeof(inq));
    memset(d, 0x7f, sizeof(d));
    memset(cnt, 0, sizeof(cnt));
    queue<int> Q;
    d[s] = 0, inq[s] = 1, Q.push(s);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        inq[u] = 0;
        for (int i = 0; i < G[u].size(); i++) {
            int v = G[u][i].X,w=G[u][i].Y;
            if (d[v] > d[u] + w) {
                d[v] = d[u] + w;
                if (!inq[v]) {
                    inq[v] = 1, Q.push(v);
                    if (++cnt[v] > n + 1) return false;
                }
            }
        }
    }
    return true;
}

void init() {
    for (int i = 0; i <= n; i++) G[i].clear();
}

int main() {
    while (scanf("%d%d", &n, &m) == 2 && n) {
        init();
        for (int i = 1; i <= n; i++) {
            G[0].push_back(mp(i, 0));
        }
        for (int i = 0; i < m; i++) {
            int cmd, a, b,c;
            scanf("%d", &cmd);
            if (cmd == 1) {
                scanf("%d%d%d", &a, &b, &c);
                G[a].push_back(mp(b, -c));
            }
            else if (cmd == 2) {
                scanf("%d%d%d", &a, &b, &c);
                G[b].push_back(mp(a, c));
            }
            else {
                scanf("%d%d", &a, &b);
                G[a].push_back(mp(b, 0));
                G[b].push_back(mp(a, 0));
            }
        }
        bool ans = spfa(0);
        if (ans) puts("Yes");
        else puts("No");
    }
    
    return 0;
}

/*
3 3
3 1 2
1 1 3 1
2 2 3 2
*/