The Mountain of Gold? UVALive - 6800

最新推荐文章于 2019-08-17 22:48:14 发布

菜得不行的sb

最新推荐文章于 2019-08-17 22:48:14 发布

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分类专栏：图

本文链接：https://blog.youkuaiyun.com/cys460714380/article/details/79801861

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本文介绍了一种用于检测是否存在从起点出发并回到起点的负权回路的算法。通过多次松弛操作，确保了能够发现图中可能存在的负权回路。

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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;
}