1035 火车停留 最大费用最大流

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<functional>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
/*
抽象到图：
找一系列首尾不相连的线段（最多n串），让这些线段的权值之和最大
最大费用最大流
将所有线段拆分为入点和出点，在入点和出点之间加边（容量为1，费用为cost[i]*0.01)，在不相邻的线段之间建边（容量为1，费用为0）
建立一个源点，向所有入点加边，容量为1，费用为0
建立一个汇点，所有出点向汇点加边，容量为1，费用为0
求出最大费用最大流
*/
const int MAXN = 1000;
const int MAXM = 10000;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to, next, cap, flow;
    double cost;
}edge[MAXM];
int head[MAXN], tol;
int pre[MAXN];
double dis[MAXN];
int beg[MAXN], ed[MAXN];
double c[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, double 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);
                    }
                }
        }
    }
    //cout << dis[t] << endl;
    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, double &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;
}
bool to[MAXN][MAXN];
int main()
{
    //源点 为2*m + 2 汇点为2 * m + 1
    ios::sync_with_stdio(0);
    int num, m;
    cin >> num >> m;
    init(2 * m + 3);
    for (int i = 1; i <= m; i++)
    {
        cin >> beg[i] >> c[i] >> ed[i];
        ed[i] += beg[i];
        c[i] *= 0.01;
        addedge(0, i, 1, 0);
        addedge(i + m, 2 * m + 1, 1, 0);
        addedge(i, i + m, 1, -c[i]);
    }
    for (int i = 1; i <= m; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (j == i) continue;
            if (ed[i] < beg[j])
                addedge(i + m, j, 1, 0);
        }
    }
    addedge(2 * m + 2, 0, num, 0);
    double ans;
    minCostMaxflow(2 * m + 2, 2 * m + 1, ans);
    printf("%.2lf\n", -ans);
}