UVA 1516 Smoking gun 查分约束+拓扑

题意：给你几个人的位置，他们可能会开枪，再给你几组关系，A B C，代表A听到B在C之前开枪，问你能否确定他们的开枪顺序。

思路：对于一组关系A B C，可以得出以下公式：fb+tb <= fc+tc，fb代表b的开火时间，tb代表b从开火到声音传到A所需的时间，因为先听到b，所以用<=，移项可得一组不等式，建边，跑一边spfa判断是否有负环，如果没有，则拓扑排序，输出解即可。

这道题我交了能有十来遍，最后发现double下0x3f3f3f3f不能初始化T T，还是自己懂得太少了。。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <string>
#include <map>
#include <set>
using namespace std;

const int maxn = 100 + 10;
const int maxe = 2000 + 10;
const double INF = 0x3f3f3f3f;
const double eps = 1e-8;

struct Edge{
int v;
double d;
int next;
Edge(int v = 0, double d = 0, int next = 0) : v(v), d(d), next(next) {}
};
struct Point{
double x, y;
int id;
Point(double x = 0, double y = 0, int id = 0) : x(x), y(y), id(id){}
void read(int _id){
scanf("%lf%lf", &x, &y);
id = _id;
}
};
int n, m, _size;
int Head[maxn], cntE;
int in[maxn];
double dist[maxn];
int cnt[maxn];
int vis[maxn];
int ans[maxn], t;
set<int> member;
map<string, Point> ma;
map<int, string> ma2;
vector<int> G[maxn];
Point p[maxn];
Edge edge[maxe];

int dcmp(double x){
if(fabs(x) < eps) return 0;
return x < 0 ? -1 : 1;
}
void init(){
for(int i = 0; i < n; i++) G[i].clear();
memset(Head, -1, sizeof(Head));
memset(in, 0, sizeof(in));
cntE = 0;
member.clear();
ma2.clear();
ma.clear();
}
void add(int u, int v, double d){
edge[cntE] = Edge(v, d, Head[u]);
Head[u] = cntE++;
}
double pow_2(double x){ return x * x; }
double Dis_2(string str1, string str2){
return sqrt(pow_2(ma[str2].x - ma[str1].x) + pow_2(ma[str2].y - ma[str1].y));
}
int spfa(int s){
for(int i = 0; i < n; i++) dist[i] = INF;
memset(vis, 0, sizeof(vis));
memset(cnt, 0, sizeof(cnt));
dist[s] = 0;
cnt[s] = 1;
queue<int> Q;
Q.push(s);
while(!Q.empty()){
int u = Q.front(); Q.pop();
vis[u] = 0;
for(int i = Head[u]; ~i; i = edge[i].next){
int v = edge[i].v;
if(dist[v] > dist[u] + edge[i].d){
dist[v] = dist[u] + edge[i].d;
if(!vis[v]){
vis[v] = 1;
Q.push(v);
if(++cnt[v] > _size) return -1;
}
}
}
}
int f = 0;
for(int i = 0; i < n; i++)if(dcmp(dist[i]) < 0){
G[i].push_back(s);
in[s]++;
f = 1;
}
return f;
}
void build(){
string str1, str2, str3, s;
for(int i = 0; i < m; i++){
/*Andy heard BillyTheKid firing before John*/
cin >> str1 >> s >> str2 >> s >> s >> str3;
int u = ma[str3].id;
int v = ma[str2].id;
double dis = Dis_2(str1, str3) - Dis_2(str1, str2);
add(u, v, dis);
member.insert(ma[str2].id);
member.insert(ma[str3].id);
}
}
void solve(){
cin >> n >> m;
init();
string str1;
for(int i = 0; i < n; i++){
cin >> str1;
p[i].read(i);
ma[str1] = p[i];
ma2[p[i].id] = str1;
}
build();
_size = (int)member.size();
for(int i = 0; i < n; i++){
int flag = spfa(i);
if(flag < 0){
cout << "IMPOSSIBLE" << endl;
return;
}
if(flag) member.erase(i);
}
t = 0;
while((int)member.size()){
if((int)member.size() > 1){
cout << "UNKNOWN" << endl;
return;
}
int cur = *member.begin();
member.erase(member.begin());
ans[t++] = cur;
for(int i = 0; i < (int)G[cur].size(); i++){
--in[G[cur][i]];
if(!in[G[cur][i]]) member.insert(G[cur][i]);
}
}
for(int i = 0; i < t; i++){
if(i) cout << " ";
cout << ma2[ans[i]];
}
cout << endl;
}
int main()
{
int T;
cin >> T;
while(T--) solve();
return 0;
}