Task Schedule

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题意：某个工厂有M台机器，需要完成N项任务。给出每项任务的完成时间PI，开始时间SI，结束时间EI；现在问你能否在刚好完成（一次）；

解析：以时间为单元构图，即将完成某一任务 I ,所需要时间PI，拆分成PI个单元1.然后设置源点s( 0 )和终点t（ n + maxday + 1 ).那么从源点到完成第i个任务则其权值为完成当前任务所需要的时间PI。然后从当前任务 i 与其完成的时段SI ~ EI 赋值为1（ 每天一台机器，所以完成的是1个单元，从而进行连通）。同理，有M台机器，因此，到达终点的权值为M，即1 * M；

然后使用最大流即可。

#include <fstream>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int MAXN = 100010;//点数最大值
const int MAXM = 400010;//边数最大值
const int INF = 0x3f3f3f3f; 

struct Edge {
    int to,next,cap,flow;
}edge[MAXM];//注意是MAXN
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN]; 

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

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;
}
/*最大流模板*/

int main(){
	int Case;
	int n, m;
	scanf( "%d", &Case );
	for( int k = 1; k <= Case; ++k ){
		scanf( "%d%d", &n, &m );
		init();
		int value, start, end, s = 0, e = 0,sum = 0, Max = 0;
		for( int i = 1; i <= n; ++i ){
			scanf( "%d%d%d", &value, &start, &end );
			Max = max( Max, end );
			sum += value;
			addedge( s, i, value );
			for( int j = start; j <= end; ++j ){
				addedge( i, n + j, 1 );
			}
		}
		int N = n + 1 + Max;
		for( int i = 1; i <= Max; ++i ){
			addedge( i + n, N, m );
		}
		int ans = sap( s, N, N);
		if( ans == sum ){
			printf( "Case %d: Yes\n\n", k );
		}
		else{
			printf( "Case %d: No\n\n", k );
		}
	}
	return 0;
}  

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