hdu 6437 Videos 最小费用最大流

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6437

因为每个视频看的人数都有限制（只能一个人看），所以可以用流量表示这个限制的值为1，然后求最大值可以把费用取反变成求最小费用。

对每个视频拆点，中间的费用为-w【i】，开头到其和其到结尾的费用都是0，保证只计算一次值。两个起点用来限制输入的人数，流量为k，其余的流量都为1。

建模真的挺重要的

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define mod 998244353
#define For(i,m,n) for(int i=m;i<=n;i++)
#define Dor(i,m,n) for(int i=m;i>=n;i--)
#define LL long long
#define lan(a,b) memset(a,b,sizeof(a))
#define sqr(a) a*a
#define type int
using namespace std;

//s源点t汇点N总点数

const int maxn=1005;
const int maxm=100005;
int n,m;
int s, t;
struct node {
    int u, v, next;
    type cap, flow, cost;
}edge[maxm];
int head[maxn], cnt;
int pre[maxn];
type dis[maxn];
bool vis[maxn];
int N;

void init () {
    memset (head, -1, sizeof head);
    cnt = 0;
}

void add_edge (int u, int v, type cap, type cost) {
    edge[cnt].u = u, edge[cnt].v = v, edge[cnt].cap = cap, edge[cnt].flow = 0;
    edge[cnt].cost = cost, edge[cnt].next = head[u], head[u] = cnt++;
    edge[cnt].u = v, edge[cnt].v = u, edge[cnt].cap = 0, edge[cnt].flow = 0;
    edge[cnt].cost = -cost, edge[cnt].next = head[v], head[v] = cnt++;
}

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

int MCMF (int s, int t, type &cost) {
    type flow = 0;
    cost = 0;
    while (spfa (s, t)) {
        type Min = INF;
        for (int i = pre[t]; i != -1; i = pre[edge[i^1].v]) {
            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].v]) {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost*Min;
        }
        flow += Min;
    }
    return flow;
}

int l[210],r[210],w[210],op[210];


int main()
{
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        init();
        int n,m,k,W;
        scanf("%d%d%d%d",&n,&m,&k,&W);
        For(i,1,m)
        {
            scanf("%d%d%d%d",&l[i],&r[i],&w[i],&op[i]);
        }
        s=0,t=2*m+2;
        add_edge(0,1,k,0);
        For(i,1,m)add_edge(1,i+1,1,0),add_edge(i+1,i+1+m,1,-w[i]),add_edge(i+1+m,t,1,0);
        For(i,1,m)
        {
            For(j,i+1,m)
            {
                int tem;
                if(op[i]==op[j])tem=W;
                else tem=0;
                if(r[i]<=l[j])
                    add_edge(i+1+m,j+1,1,tem);
                else if(r[j]<=l[i])
                    add_edge(j+1+m,i+1,1,tem);
            }
        }
        int ans;
        N=2*m+3;
        MCMF(s,t,ans);
        printf("%d\n",-ans);
    }
    return 0;
}