[HDU 3572]Task Schedule[最大流]

题目链接：[HDU 3572]Task Schedule[最大流]

题意分析：

N个任务，每个只能在si之后开始执行，得执行pi天，在ei天前得执行完，有M个机器进行作业，问：能否在每个的规定时间之内，把所有任务执行完？任务可以中断执行，比如第一天做，第二天休息，第三天做。

解题思路：

我们将每天视为一个点，那么第i个任务可以在si到ei之间执行，那么说明这个任务可以和[si,ei]之间的点连一天流量为1的边，代表被执行了一天，接着我们将源点与第i个任务连一条容量为pi的边，代表这个任务要被执行的次数，最后我们让每天和汇点之间连一条容量为m的边，表示这天最多执行m个任务，如果最大流等于需要被执行的总天数，则Yes，否则No。

个人感受：

哇(⊙０⊙)，神奇的建图。网络流老想着贪心这个梗估计能接着陪完玩到毕业啊= =。

具体代码如下：

#include<algorithm>#include<cctype>#include<cmath>#include<cstdio>#include<cstring>#include<iomanip>#include<iostream>#include<map>#include<queue>#include<set>#include<sstream>#include<stack>#include<string>#define lowbit(x) (x & (-x))#define root 1, n, 1#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1#define ll long long#define pr(x) cout << #x << " = " << (x) << '\n';using namespace std;const int MAXN = 1010;//点数的最大值const int MAXM = 1e6;//边数的最大值const int INF = 0x3f3f3f3f;struct Edge{    int to,next,cap,flow;}edge[MAXM];//注意是MAXMint tol;int head[MAXN];int gap[MAXN],dep[MAXN],cur[MAXN];void addedge(int u,int v,int w,int rw = 0){    edge[tol].to = v; edge[tol].cap = w; edge[tol].flow = 0;    edge[tol].next = head[u]; head[u] = tol++;    edge[tol].to = u; edge[tol].cap = rw; edge[tol].flow = 0;    edge[tol].next = head[v]; head[v] = tol++;}int Q[MAXN];void BFS(int start,int end){    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0] = 1;    int front = 0, rear = 0;    dep[end] = 0;    Q[rear++] = end;    while(front != rear)    {        int u = Q[front++];        for(int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].to;            if(dep[v] != -1)continue;            Q[rear++] = v;            dep[v] = dep[u] + 1;            gap[dep[v]]++;        }    }}int S[MAXN];int sap(int start,int end,int N){    BFS(start,end);    memcpy(cur,head,sizeof(head));    int top = 0;    int u = start;    int ans = 0;    while(dep[start] < N)    {        if(u == end)        {            int Min = INF;            int inser;            for(int i = 0;i < top;i++)            if(Min > edge[S[i]].cap - edge[S[i]].flow)            {                Min = edge[S[i]].cap - edge[S[i]].flow;                inser = i;            }            for(int i = 0;i < top;i++)            {                edge[S[i]].flow += Min;                edge[S[i]^1].flow -= Min;            }            ans += Min;            top = inser;            u = edge[S[top]^1].to;            continue;        }        bool flag = false;        int v;        for(int i = cur[u]; i != -1; i = edge[i].next)        {            v = edge[i].to;            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])            {                flag = true;                cur[u] = i;                break;            }        }        if(flag)        {            S[top++] = cur[u];            u = v;            continue;        }        int Min = N;        for(int i = head[u]; i != -1; i = edge[i].next)            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)            {                Min = dep[edge[i].to];                cur[u] = i;            }        gap[dep[u]]--;        if(!gap[dep[u]]) return ans;        dep[u] = Min + 1;        gap[dep[u]]++;        if(u != start)u = edge[S[--top]^1].to;    }    return ans;}void init() {    tol = 0;    memset(head,-1,sizeof(head));}int main(){    #ifdef LOCAL    freopen("C:\\Users\\apple\\Desktop\\in.txt", "r", stdin);    #endif    int n, m;    for (int kk, kase = scanf("%d", &kk); kase <= kk; ++kase) {        init();        scanf("%d%d", &n, &m);        int s, p, e;        int src = 0, des = 0, mx = 0, sum = 0;        for (int i = 1; i <= n; ++i) {            scanf("%d%d%d", &p, &s, &e);            addedge(src, i, p);            sum += p;            mx = max(mx, e);            for (int j = s; j <= e; ++j) {                addedge(i, j + n, 1);            }        }        des = mx + n + 1;        for (int i = 1; i <= mx; ++i) {            addedge(i + n, des, m);        }        printf("Case %d: ", kase);        if (sap(src, des, des + 1) == sum) printf("Yes\n\n");        else printf("No\n\n");    }    return 0;}