Luo Gu P2294 [HNOI2005] cunning businessman

Luo Gu P2294 [HNOI2005] cunning businessman

topic:

• There are pictures·. Turn link

answer:

• Differential constraints.
• Although the title does not appear inequality, but still fall within the scope of the differential constraints.
• Start I in accordance with the requirements of its edge from u to v in the weight w. But I found not. So I added up another u to v -w is right side, then on the line. The Discovery of the differential constraint problem did not get out often there is no constraint to find full .

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 100005
using namespace std;

struct E {int next, to, dis;} e[N];
int T, n, m, num, flag;
int h[N], dis[N], cnt[N];
bool vis[N];

void add(int u, int v, int w)
{
    e[++num].next = h[u];
    e[num].to = v;
    e[num].dis = w;
    h[u] = num;
}

int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return x *= f;
}

bool spfa(int s)
{
    queue<int> que;
    memset(dis, -0x3f, sizeof(dis));
    dis[s] = 0, vis[s] = 1, que.push(s);
    while(!que.empty())
    {
        int now = que.front();
        que.pop();  vis[now] = 0;
        cnt[now]++; if(cnt[now] == n) return 0;
        for(int i = h[now]; i != 0; i = e[i].next)
            if(dis[now] + e[i].dis > dis[e[i].to])
            {
                dis[e[i].to] = dis[now] + e[i].dis;
                if(!vis[e[i].to])
                    vis[e[i].to] = 1, que.push(e[i].to);
            }
    }
    return 1;
}

int main()
{
    cin >> T;
    while(T--)
    {
        flag = num = 0;
        memset(h, 0, sizeof(h));
        memset(cnt, 0, sizeof(cnt));
        memset(vis, 0, sizeof(vis));
        n = read(), m = read();
        for(int i = 1; i <= m; i++)
        {
            int u = read(), v = read(), w = read();
            add(u - 1, v, w), add(v, u - 1, -w);
        }
        for(int i = 0; i <= n; i++) add(n + 1, i, 0);
        if(!spfa(n + 1)) printf("false\n");
        else printf("true\n");
    }
    return 0;
}