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
<< endl;
if(flag) {
cout << "-1" << endl;
}
else if(dis
== inf) {
cout << "-2" << endl;
}
else cout << dis
<< 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();
}
}