Codeforces 863F Almost Permutation

题

　　OvO http://codeforces.com/problemset/problem/863/F

　　（Educational Codeforces Round 29 - F）

解

　　费用流

　　对每个ax，从源点连一条费用为0，流量为1的边，

　　对于每个ax，对能取到的每个值，连一条费用为0，流量为1的边，

　　对于每个能取到的值（n个），向汇点连n条流量为1的边，满足：第一条到第i条边的费用和为i^2（即费用分别为1,3,5,7,9……）

代码

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <string>
#include <queue>

using namespace std;

const int M=55;
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int to,next,cap,flow,cost;
}edge[MAXM];

int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1

void init(int n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int cap,int cost)
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}

bool spfa(int s,int t)
{
    queue<int>q;
    for(int i = 0;i < N;i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1;i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow &&
               dis[v] > dis[u] + edge[i].cost )
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t] == -1)return false;
    else return true;
}

//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
    int flow = 0;
    cost = 0;
    while(spfa(s,t))
    {
        int Min = INF;
        for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
        {
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}

int n,q;
int lmt[M][3];

void solve()
{
	int i,j,ans,lst,now,tmp;
	int s=0,t=2*n+1;
	init(2*n+2);
	for(i=1;i<=n;i++)
		if(lmt[i][0]>lmt[i][1])
		{
			printf("-1
");
			return ;
		}
	for(i=1;i<=n;i++)
		addedge(s,i,1,0);
	for(i=1;i<=n;i++)
		for(j=lmt[i][0];j<=lmt[i][1];j++)
			addedge(i,n+j,1,0);
	for(i=1;i<=n;i++)
	{
		lst=0;
		for(j=1;j<=n;j++)
		{
			addedge(n+i,t,1,j*j-lst);
			lst=j*j;
		}
	}
	minCostMaxflow(s,t,ans);
	printf("%d
",ans);
}

int main()
{
	int i,j,t,li,ri,v;
	scanf("%d%d",&n,&q);
	for(i=1;i<=n;i++)
		lmt[i][0]=1,lmt[i][1]=n;
	for(i=1;i<=q;i++)
	{
		scanf("%d%d%d%d",&t,&li,&ri,&v);
		if(t==1)
			for(j=li;j<=ri;j++)
				lmt[j][0]=max(lmt[j][0],v);
		else
			for(j=li;j<=ri;j++)
				lmt[j][1]=min(lmt[j][1],v);
	}
	solve();
	return 0;
}