差分约束系统·SPFA模板·POJ 3169·Layout

题目大意：略；

解题思路：

由于满足不等式关系，同时也满足差分约束系统，因此，我们可以通过转化为最短路来求解；

a - b <= c;

我们就可以直接add(a,b,c)建边；

而a - b >= c的那种。我们需要左右同乘以-1转化为：

b - a <= -c 也满足最短路松弛的形式；

因此add(b, a, -c)建边即可；

因为有负边，因此，使用优化的BELLMAN-ford即SPFA来进行计算

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <fstream>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define   maxn          1010
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   ms(x,y)      memset(x,y,sizeof(x))
#define   rep(i,n)      for(int i=0;i<(n);i++)
#define   repf(i,a,b)   for(int i=(a);i<=(b);i++)
#define   pii           pair<int,int>
//#define   mp            make_pair
#define   FI            first
#define   SE            second
#define   IT            iterator
#define   PB            push_back
#define   Times         10
typedef   long long     ll;
typedef   unsigned long long ull;
typedef   long double   ld;
typedef   pair<int ,int > P;

const double eps = 1e-10;
const double  pi = acos(-1.0);
const  ll    mod = 1e9+7;
const  int   inf = 0x3f3f3f3f;
const  ll    INF = (ll)1e18+300;
const int    maxd = 200000;

struct node {
    int from;
    int to;
    int w;
    int next;
}e[maxd];
int head[10000];
int vis[10000];
int out[10000];
int dis[10000];
int n, m1, d1, cont;

void add(int from, int to, int w) {
    //e[cont].from = from;
    e[cont].to = to;
    e[cont].w = w;
    e[cont].next = head[from];
    head[from] = cont ++;
}
void SPFA() {
    for (int i = 1; i <= n; i++) {
        dis[i] = inf;
    }
    dis[1] = 0;
    ms(vis, 0);
    ms(out, 0);
    queue<int > que;
    que.push(1);
    int flag = 0;
    while(!que.empty()) {
        int u = que.front();
        out[u] ++;
        if(out[u] > n) {
            flag = 1;
            break;
        }
        que.pop();
        vis[u] = 0;
        for (int i = head[u]; i != -1; i = e[i].next) {
            int v = e[i].to;
            int w = e[i].w;
            if(dis[v] > dis[u] + w) {
                dis[v] = dis[u] + w;
                if(!vis[v]) {
                    vis[v] = 1;
                    que.push(v);
                }
            }
        }
    }
    //cout << dis[n] << endl;
    if(flag) {
        cout << "-1" << endl;
    }
    else if(dis[n] == inf) {
        cout << "-2" << endl;
    }
    else cout << dis[n] << endl;
}

int main() {
    while (~scanf("%d%d%d", &n, &m1, &d1)) {
        cont = 0;
        ms(head, -1);
        for (int i = 0; i < m1; i++){
            int x, y, w;
            scanf("%d%d%d",&x, &y, &w);
            add(x, y, w);
        }
        for (int i = 0; i < d1; i++) {
            int x, y, w;
            scanf("%d%d%d", &x, &y, &w);
            add(y, x, -w);
        }
        SPFA();
    }
}