POJ 3169 Layout (差分约束)

解题思路：很标准的差分约束题目，根据题意列出不等式，并将所有的不等式转化为 ≤ 形式，然后跑最短路即可。

AC代码：

/** @Author: wchhlbt* @Last Modified time: 2017-09-19*/#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>#include <limits>#include <climits>#include <cstdio>#define Fori(x) for(int i=0;i<x;i++)#define Forj(x) for(int j=0;j<x;j++)#define maxn 10007#define inf 0x3f3f3f3f#define ONES(x) __builtin_popcount(x)#define pb push_back#define AA first#define BB second#define _  << "  " <<using namespace std;typedef long long ll ;const double eps =1e-8;const int mod = 998244353;const double PI = acos(-1.0);int dx[5] = {0,0,1,-1,0};int dy[5] = {1,-1,0,0,0};inline int read(){ int cnt;    scanf("%d",&cnt);   return cnt;}//每次使用前需要调用init函数初始化 可以处理负权边//最坏复杂度O(V*E)int d[maxn],inq[maxn];//inq数组储存当前点是否在队列中int deg[maxn];//记录每个节点入队次数vector<pair<int,int> > e[maxn]; //pair<节点, 边权> int n;void init(){for(int i = 0; i<maxn; i++){e[i].clear();d[i] = inf;inq[i] = 0;deg[i] = 0;}}int SPFA(int s)//s为起点{queue<int> Q;Q.push(s); d[s] = 0; inq[s] = 1; deg[s] = 1;while(!Q.empty()){int hd = Q.front();Q.pop();inq[hd] = 0;for(int i = 0; i<e[hd].size(); i++){int u = e[hd][i].first;int v = e[hd][i].second;if(d[u]>d[hd]+v){d[u] = d[hd] + v;if(inq[u]==1)continue;inq[u] = 1;deg[u]++;if(deg[u]>=n)return -1;Q.push(u);}}}if(d[n]==inf)return -2;return d[n];}int main(){int x,y;while(~scanf("%d%d%d",&n,&x,&y)){init();for(int i = 0; i<x; i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);e[a].pb(make_pair(b,c));//e[b].pb(make_pair(a,c));}for(int i = 0; i<y; i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);//e[a].pb(make_pair(b,-c));e[b].pb(make_pair(a,-c));}cout << SPFA(1) << endl;}    return 0;}