uva11478（差分约束，spfa求负环）

UVA - 11478 Halum （最短路应用+二分）

Description

	Problem H	Halum	Time Limit : 3 seconds
	You are given a directed graph G(V,E) with a set of vertices and edges. Each edge (i,j) that connects some vertex i to vertex j has an integer cost associated with that edge.   Define the operation Halum(v, d) to operate on a vertex v using an integer d as follows: subtract d from the cost of all edges that enter v and add d to the cost of every edge that leaves v. As an example of that operation, consider graph G that has three vertices named (1, 2, 3) and two edges. Edge (1, 2) has cost -1, and edge (2,3) has cost 1. The operation Halum(2,-3) operates on edges entering and leaving vertex 2.  Thus, edge (1, 2) gets cost -1-(-3)=2 and the edge (2, 3) gets cost 1 + (-3) = -2. Your goal is to apply the Halum function to a graph, potentially repeatedly, until every edge in the graph has at least a certain cost that is greater than zero. You have to maximize this cost.
	Input
	Two space-separated integers per case: V(V≤500) and E(E≤2700). E lines follow. Each line represents a directed edge using three space-separated integers (u, v, d). Absolute value of cost can be at most 10000.
	Output
	If the problem is solvable, then print the maximum possible value. If there is no such solution print “No Solution”. If the value can be arbitrary large print “Infinite”
	Sample Input	Sample Output
	2 1 1 2 10 2 1 1 2 -10 3 3 1 2 4 2 3 2 3 1 5 4 5 2 3 4 4 2 5 3 4 2 3 1 0 1 2 -1	Infinite Infinite 3 1

题意：给定一个有向图，每条边都有一个权值，每次你可选择一个结点v和一个整数d，把所有以v为终点的边的权值减少d，把所有以v为终点的边的权值增加d，最后要让所有边的最小值非负且尽量大。

思路：最小值最大，大牛们想到用二分求答案，令sum(u)表示为作用在节点u之上的所有d之和，这样题目的目标就是确定所有的sum(u)了； 对于边a->b，不难发现操作后的权值是:w(a, b)+sum(a)-sum(b)>=x

那么就可以得到个不等式：sum(b)-sum(a) <= w(a, b)-x，而w(a, b)-x我们是已知的，那么就能得到相当于最短路的

不等式 d[v]<=d[u]+w(u, v)，这样，我们实际得到一个差分约束系统。，那么我们在做spfa的时候，如果发现负权环的话，那么就相当于我们无法得到一个类似最短路的不等式，所以无解。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
const int N=505;
typedef long long ll;
int dis[N];
int cnt[N],m,n;
bool vis[N];
int head[N];
int ip;
struct edgenode
{
    int to,w;
    int next;
} tu[N*N];

void add(int u,int v,int w)
{
    tu[ip].to=v,tu[ip].w=w,tu[ip].next=head[u],head[u]=ip++;
}
bool spfa()
{
    queue<int>q;
    for(int i=1; i<=n; i++)
        dis[i]=0,cnt[i]=0,vis[i]=0,q.push(i);
    while(!q.empty())
    {
        int t=q.front();
        q.pop();
        vis[t]=0;
        for(int k=head[t]; k!=-1; k=tu[k].next)
            if(dis[t]+tu[k].w<dis[tu[k].to])
            {
                dis[tu[k].to]=dis[t]+tu[k].w;
                if(!vis[tu[k].to])
                {
                    vis[tu[k].to]=1;
                    q.push(tu[k].to);
                    if(++cnt[tu[k].to]>=n)
                        return 0;
                }
            }
    }
    return 1;
}

bool judge(double x)
{
    bool flag=0;
    for(int i=1; i<=n; i++)
        for(int k=head[i]; k!=-1; k=tu[k].next)
            tu[k].w-=x;
    if(!spfa())flag=1;
    for(int i=1; i<=n; i++)
        for(int k=head[i]; k!=-1; k=tu[k].next)
            tu[k].w+=x;
    return flag;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        ip=0;
        memset(head,-1,sizeof(head));
        int l=1,r=0,mid;
        while(m--)
        {
            int a,b,w;
            scanf("%d%d%d",&a,&b,&w);
            add(a,b,w);
            r=max(r,w);
        }
        if(!judge(r))
            printf("Infinite\n");
        else if(judge(l))
            printf("No Solution\n");
        else
        {
            int ans=l++;
            while(l<r)
            {
                mid=(r+l)/2;
                if(judge(mid))
                    r=mid;
                else
                    l=mid+1,ans=mid;
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}