题意 给出N个任务和M台机器

任务中给出 C  S E 分别表示 所需时间  任务开始时间  任务结束时间

即任务需要在开始到结束这时间段中完成。

一个任务可以在不同的机器上完成，但每次在机器上至少要运行一天

建边， 源点-任务 容量为C

任务-时间段上的每一个时间点 容量为1

时间点-汇点 容量为 M

最大流 >= sigma(C)则YES else NO

（最大所需点数为n+m+2;)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <math.h>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <time.h>;
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define IN     freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 200010;//点数的最大值
const int MAXM = 200006;//边数的最大值
const int INF = 0x3f3f3f3f;
const int mod = 10000007;
struct Edge
{int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
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;int i;while(dep[start] < N){if(u == end){int Min = INF;int inser;for( 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( 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( 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( 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;
}
int main()
{//IN;int t,cas=0,m,n;scanf("%d",&t);while(t--){init();scanf("%d %d",&n,&m);int e=500+n+1,sum=0;for(int i=1;i<=500;i++){addedge(n+i,e,m);}for(int i=1;i<=n;i++){int x,y,c;scanf("%d %d %d",&c,&x,&y);sum+=c;addedge(0,i,c);//源点到任务for(int j=x;j<=y;j++)//任务到天{addedge(i,n+j,1);}}printf("Case %d: ",++cas);int out=sap(0,e,n+500+2);//  printf("%d\n",out);if(out>=sum)puts("Yes");else puts("No");puts("");}
}