差分约束 刷题记录

把问题转化成一堆不等式，然后用最短路求解

POJ3169 Layout

最后要求1和n之间最大dis是多少  ->  转化为得到一堆  d[n] - d[1] <= xi  然后求xi的最小值

对于给出的是d[u] - d[v] >= xi  同乘-1转化为 d[v] - d[u] <= -xi 即可

然后用spfa求 1到n的最短路

若有负环则无解，若求得的最短路为inf则说明1,n间距离可随便大

代码：

#include <cstdio>
#include <iostream>
#include <queue>
#define nmax 50010
#define f(c,a,b) for(int c=a; c<=b; c++)

using namespace std;
typedef long long ll;
struct edge{
    int v, ne, w;
}e[nmax];
int cnt=0, n, x, y;
int head[nmax]={0}, c[nmax]={0}, inq[nmax]={0};
ll d[nmax]={0};
const ll inf = 1e15;

void addedge(int u, int v, int w){
    cnt++;
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].ne = head[u];
    head[u] = cnt;
}

void build(){
    scanf("%d%d%d", &n, &x, &y);
    int u, v, w;
    f(i, 1, x){
        scanf("%d%d%d", &u, &v, &w);
        addedge(u, v, w);
    }
    f(i, 1, y){
        scanf("%d%d%d", &u, &v, &w);
        addedge(v, u, -w);
    }
    f(i, 2, n) d[i] = inf;
}

bool spfa(){
    queue <int> q;
    q.push(1);
    inq[1] = c[1] = 1;
    while(!q.empty()){
        int u = q.front();
        if( c[u] > n ) return false;
        q.pop();
        inq[u] = 0;
        for (int i=head[u]; i; i=e[i].ne){
            int v = e[i].v;
            if(d[v] > d[u] + e[i].w){
                d[v] = d[u] + e[i].w;
                if(inq[v]) continue;
                inq[v] = 1;
                c[v]++;
                q.push(v);
            }
        }
    }
    return true;
}

int main(){
    build();
    if( spfa() ){
        if(d[n]==inf) printf("-2\n");
        else printf("%lld\n", d[n]);
    }else{
        printf("-1\n");
    }
    return 0;
}