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The Mountain of Gold? UVALive - 6800

2018-04-03 13:16 363 查看
https://cn.vjudge.net/problem/UVALive-6800

题意:

问是否有从0出发并且回到0的负环

假如没有负环每个点最多被松弛n次

所以直接对每条边松弛n次

再看能不能松弛

可以的话就说明存在这样的负环

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const ll mod = 1000000007 ;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 100000000000000000;//1e17
const int maxn=  2e3+20;
const int maxm = 3e2+20;
//muv[i]=(p-(p/i))*muv[p%i]%p;
int in(int &ret) {
char c;
int sgn ;
if(c=getchar(),c==EOF)return -1;
while(c!='-'&&(c<'0'||c>'9'))c=getchar();
sgn = (c=='-')?-1:1;
ret = (c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
ret *=sgn;
return 1;
}

int dp[maxn];
bool vis[maxn];
vector<pii>g[maxn];
int a[maxn],b[maxn],c[maxn];
int dis[maxn];
bool dfs(int u) {
if(u==0)return 1;
if(vis[u])return dp[u];
vis[u] = 1;
int sz = g[u].size();
r0(i,sz) {
int v = g[u][i].X;
if(dfs(v))return dp[u] = 1;
}
return dp[u] = 0;
}
int main() {
#ifdef LOCAL
freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL
int t;
sd(t);
r1(cas,t) {
printf("Case #%d: ",cas);
int n,m;
sdd(n,m);
r0(i,n)g[i].clear(),vis[i] = 0,dis[i] = inf,dp[i] = 0;
r1(i,m) {
int u,v,w;
sddd(u,v,w);
a[i] = u,b[i] = v,c[i] = w;;
g[u].pb(mp(v,w));
}
dis[0] = 0;
r0(i,n) {
r1(j,m) {
int u = a[j],v = b[j],w = c[j];
if(dis[v]>dis[u]+w) {
dis[v] = dis[u] + w;
}
}
}
bool ok = 0;
for(int i=1; i<=m&&!ok; ++i) {
int u = a[i],v = b[i];
if(dis[v]>dis[u] + c[i]) {
if(dfs(u))ok = 1;
}
}
puts(ok?"possible":"not possible");
}
return 0;
}
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