POJ-2983 Is the Information Reliable?

题目大意：

有n个点，有m个约束，有两种约束形式，一种是P A B C表示A在B的北边距离为C的地方，另外一种是V A B表示A在B的背边距离至少为1的地方，问你这个信息是否存在矛盾的地方

解题思路：

P A B C表示S[B] - S[A] = C

那么可以表示成C <= S[B] - S[A]  <= C

这样就是差分约束的模板题了

代码：

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

const int maxn = 1000 + 5;
const int INF = 0x3f3f3f3f;

typedef struct node{
    int to, w;
    int next;
    node(int a = 0, int b = 0, int c = 0){
        to = a; next = b; w = c;
    }
}Edge;

int tot;
//int s[maxn * maxn];
Edge edge[maxn * maxn];
int head[maxn * maxn], dis[maxn], vis[maxn], cnt[maxn];

void add(int u, int v, int w){
    edge[tot] = node(v, head[u], w);
    head[u] = tot++;
}
bool spfa(int n){
    int u, v;// top = 0;
    queue<int> q;
    while(!q.empty()) q.pop();
    for(int i = 0; i <= n; ++i){
        dis[i] = INF;
        vis[i] = 0; cnt[i] = 0;
    }
    //s[top++] = 0;
    q.push(0);
    dis[0] = 0; vis[0] = 1;
    while(!q.empty()){
        u = q.front(); q.pop(); vis[u] = 0;
        if((++cnt[u]) > n) return 0;
        for(int i = head[u]; ~i; i = edge[i].next){
            v = edge[i].to;
            if(dis[v] > dis[u] + edge[i].w){
                dis[v] = dis[u] + edge[i].w;
                if(!vis[v]){
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return 1;
}
int main(){
    char op;
    int a, b, x, n, m;
    while(~scanf("%d%d", &n, &m)){
        tot = 0;
        memset(head, -1, sizeof(head));
        for(int i = 1; i <= n; ++i) add(0, i, 0);
        for(int i = 0; i < m; ++i){
            scanf(" %c%d%d", &op, &a, &b);
            if(op == 'P'){
                scanf("%d", &x);
                add(a, b, -x);
                add(b, a, x);
            }else add(a, b, -1);
        }
        if(spfa(n)) puts("Reliable");
        else puts("Unreliable");
    }
    return 0;
}