UVA11478 Halum

来源：互联网发布：广点通优化思路编辑：程序博客网时间：2024/05/21 17:37

2分法，对于最后能否达到x应满足每一对的：
edge(u, v) + sum(u) - sum(v) >= x
-->sum(v) <= sum(u) + (edge(u, v) - x)
-->d[v] <= d[u] + edge0(u, v)
-->模拟建图，判断是否有负环

注意:题目的目标是所有边的权值都大于0而不是非负

#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>#include <queue>using namespace std;const int maxn = 500+5;class Edge{    public:    int pos, dist;    Edge(int pos, int dist)    {        this->pos = pos;        this->dist = dist;    }};vector<Edge> adj[maxn];bool inq[maxn];int d[maxn], cnt[maxn], n, m, up;void Init(){    for(int i = 1; i <= n; i++)        adj[i].clear();    int a, b, c;    up = 0;    for(int i = 1; i <= m; i++)    {        scanf("%d%d%d", &a, &b, &c);        adj[a].push_back(Edge(b, c));        up = max(up, c);    }}bool Bellman_Ford(){    queue<int> myQue;    for(int i = 1; i <= n; i++)    {        myQue.push(i);        inq[i] = true;        d[i] = 0;        cnt[i] = 0;    }    while(!myQue.empty())    {        int u = myQue.front();        myQue.pop();        inq[u] = false;        for(vector<Edge>::iterator it = adj[u].begin(); it != adj[u].end(); it++)        {            int v = it->pos;            if(d[v] > d[u] + it->dist)            {                d[v] = d[u] + it->dist;                if(!inq[v])                {                    myQue.push(v);                    inq[v] = true;                    if(++cnt[v] > n)                        return true;//有负环                }            }        }    }    return false;}bool Test(int M){    for(int i = 1; i <= n; i++)        for(vector<Edge>::iterator it = adj[i].begin(); it != adj[i].end(); it++)            it->dist -= M;    bool flag = Bellman_Ford();    for(int i = 1; i <= n; i++)        for(vector<Edge>::iterator it = adj[i].begin(); it != adj[i].end(); it++)            it->dist += M;    return flag;}void Solve(){    if(!Test(up+1))        printf("Infinite\n");    else if(Test(1))        printf("No Solution\n");    else    {        int L = 1, R = up + 1;        while(L < R - 1)        {            int M = L + (R - L) / 2;            if(!Test(M))                L = M;            else                R = M;        }        printf("%d\n", L);    }}int main(){    while(scanf("%d%d", &n, &m) != EOF)    {        Init();        Solve();    }    return 0;}