hdoj 3572 Task Schedule

题目链接：​​Task Schedule​​

题目大意：有n个任务和m个机器，一个任务需要ti的时间，只能从第s天开始，必须在第e天之前结束，问能不能把所有的任务都做完

题目思路：我们可以想到去建网络流，建一个超级源点0，然后第i个任务为点i，连接他和源点，容量为任务需要的天数（流出去的量嘛），然后每个任务和他的天数集合（从起点到终点）相连，容量为1，然后所有的天数和总计汇点连接，容量是m（最多只有m个机器嘛，同一天最多m个机器做工作），然后跑一遍最大流就好了

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

const int MAXN = 1000+100;
const int MAXM = 1000000+10;
const int INF = 1000000+10;

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

int dist[MAXN], vis[MAXN], cur[MAXN], top, head[MAXN];
int n, m;
int sumflow;
int sink;

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

void addedge(int u, int v, int w){
    Edge E1 = {u, v, w, 0, head[u]};
    edge[top] = E1;
    head[u] = top++;
    Edge E2 = {v, u, 0, 0, head[v]};
    edge[top] = E2;
    head[v] = top++;
}

bool BFS(int start, int End){
    queue<int> Q;
    memset(dist, -1, sizeof(dist));
    memset(vis, 0, sizeof(vis));
    while(!Q.empty()) Q.pop();
    Q.push(start);
    vis[start] = 1;
    dist[start] = 0;
    while(!Q.empty()){
        int u = Q.front();
        Q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next){
            Edge E = edge[i];
            if(!vis[E.to] && E.cap > E.flow){
                vis[E.to] = 1;
                dist[E.to] = dist[u] + 1;
                if(E.to == End) return true;
                Q.push(E.to);
            }
        }
    }
    return false;
}

int DFS(int x, int a, int End){
    if(x == End || a == 0) return a;
    int flow = 0, f;
    for(int& i = cur[x]; i != -1; i = edge[i].next){
        Edge& E = edge[i];
        if(dist[E.to] == dist[x]+1 && (f = DFS(E.to, min(a, E.cap-E.flow), End)) > 0){
            E.flow += f;
            edge[i^1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}

int Dinic(int start, int End){
    int flow = 0;
    while(BFS(start, End)){
        memcpy(cur, head, sizeof(head));
        flow += DFS(start, INF, End);
    }
    return flow;
}

int main(){
    int T,n,m;
    scanf("%d",&T);
    for(int Case = 1;Case <= T;Case++){
        init();
        scanf("%d%d",&n,&m);
        int ti,s, e;
        int last = 0;
        sumflow = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d%d%d", &ti, &s, &e);
            sumflow += ti;
            addedge(0, i, ti);
            last = max(last, e);
            for(int j = s; j <= e; j++)
            addedge(i, n+j, 1);
        }
        sink = n+last+1;
        for(int j = 1; j <= last; j++)
            addedge(n+j, sink, m);
        if(Dinic(0,sink) == sumflow) printf("Case %d: Yes\n\n",Case);
        else printf("Case %d: No\n\n",Case);
    }
    return 0;
}