UVA11248_Frequency Hopping

给一个有向网络，求其1,n两点的最大流量是否不小于C，如果小于，是否可以通过修改一条边的容量使得最大流量不小于C？

首先对于给定的网络，我们可以先跑一遍最大流，然后先看流量是否大于C。

然后保存跑完第一次最大流后的残余网络容量情况，然后接下来对于每条割，将分别其容量加C-maxflow,看看能否得到满足条件的流量即可。

 

 

召唤代码君：

 

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>
#define maxn 1110
#define maxm 222220
#define mp(x,y) make_pair(x,y)
using namespace std;

typedef pair<int,int> pr;
int C[maxm];
int to[maxm],next[maxm],c[maxm],first[maxn],edge;
int d[maxn],tag[maxn],TAG=222;
bool can[maxn];
int Q[maxn],bot,top;
int n,m,cmax,s,t,cas=0;
vector<pr> ans;

void _init()
{
    edge=-1;
    s=1,t=n;
    for (int i=1; i<=n; i++) first[i]=-1;
}

void addedge(int U,int V,int W)
{
    edge++;
    to[edge]=V,c[edge]=W,next[edge]=first[U],first[U]=edge;
    edge++;
    to[edge]=U,c[edge]=0,next[edge]=first[V],first[V]=edge;
}

bool bfs()
{
    Q[bot=top=1]=t,d[t]=0,tag[t]=++TAG,can[t]=false;
    while (bot<=top)
    {
        int cur=Q[bot++];
        for (int i=first[cur]; i!=-1; i=next[i])
            if (c[i^1]>0 && tag[to[i]]!=TAG)
            {
                d[to[i]]=d[cur]+1,Q[++top]=to[i];
                can[to[i]]=false,tag[to[i]]=TAG;
                if (to[i]==s) return true;
            }
    }
    return false;
}

int dfs(int cur,int num)
{
    if (cur==t) return num;
    int tmp=num,k;
    for (int i=first[cur]; i!=-1; i=next[i])
        if (c[i]>0 && tag[to[i]]==TAG && d[to[i]]==d[cur]-1 && !can[to[i]])
        {
            k=dfs(to[i],min(num,c[i]));
            if (k) num-=k,c[i]-=k,c[i^1]+=k;
            if (!num) break;
        }
    if (num) can[cur]=true;
    return tmp-num;
}

int maxflow()
{
    int flow=0;
    while (bfs()) flow+=dfs(s,cmax);
    return flow;
}

int main()
{
    int U,V,W,last;
    while (scanf("%d%d%d",&n,&m,&cmax) && (n))
    {
        _init();
        for (int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&U,&V,&W);
            addedge(U,V,W);
        }
        printf("Case %d: ",++cas);
        last=maxflow();
        if (last>=cmax)
        {
            puts("possible");
            continue;
        }
        ans.clear();
        for (int i=0; i<=edge; i++) C[i]=c[i];
        for (int i=0; i<=edge; i+=2)
        {
            if (c[i]+c[i+1]>cmax) continue;
            c[i]+=cmax-last-c[i];
            if (last+maxflow()>=cmax) ans.push_back(mp(to[i+1],to[i]));
            for (int j=0; j<=edge; j++) c[j]=C[j];
        }
        if (ans.size()==0) puts("not possible");
        else
        {
            sort(ans.begin(),ans.end());
            printf("possible option:(%d,%d)",ans[0].first,ans[0].second);
            for (unsigned i=1; i<ans.size(); i++)
                printf(",(%d,%d)",ans[i].first,ans[i].second);
            printf("\n");
        }
    }
    return 0;
}