2017 ICPCECIC 北方邀请赛 H MJF wants to work (贪心)

MJF wants to work

时间限制: 1 Sec 内存限制: 128 MB
提交: 87 解决: 16
[提交][状态][讨论版]

题目描述

MJF feel the summer vacation is very leisure. So he wants to work to earn money.
There are n jobs MJF can take part in.For job i, it has the start time ti.x, the end time ti.y and reward ci.
MJF wants to take part in two jobs (exact two jobs), and he wants to earn exactly M yuan. He doesn't want to earn one money more,he thinks it's a waste.
Becase MJF is a lazy boy, so he wants the time of the sum of two jobs shortest.
Becase MJF can only take part in one job in a time.So two jobs can't overlap.(ti.y < tj.x)

输入

The input consists of multiple test cases.
For each test，the first line contains 2 integers n, m.(1 <= n, m <= 2e5)
Then following n lines, each line contains 3 integers ti.x, ti.y, ci.(1 <= ti.x, ti.y, ci <= 2e5)

输出

For each test case, print the value of the sum of two jobs' time.if there are no answer,please print"oh no!"

样例输入

3 10

1 2 3

3 4 7

4 6 7

1 10

1 10 10

样例输出

oh no!

123

思路为:

把时间段从线段化成左右两个端点问题, 每个端点存时间, 金钱和区分左右标志,

然后按照左端点从小到大排序, 并且是一左一右排序, 遇到左端点, 维护ans 值, dp数组存的是时间, 维护 m- 当前已挣得钱的时间尽可能小

遇到右端点维护 dp 时间

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};int n,m;
int cont;
int dp[maxn*2];// dp数组存的是时间
struct node{int x;int time;int mo;int flag;//0 左端点, 1 右端点
}a[maxn*2];
int cmp(node a,node b)
{if(a.x==b.x)return a.flag<b.flag;return a.x<b.x;}
void init()
{for(int i=0;i<=maxn;i++)dp[i]=INF;int x,y,z;mem(a,0);cont=0;for(int i=1;i<=n;i++){scanf("%d %d %d",&x,&y,&z);a[++cont].x=x;a[cont].time=y-x+1;a[cont].flag=0;a[cont].mo=z;a[++cont].x=y;a[cont].time=y-x+1;a[cont].flag=1;a[cont].mo=z;}sort(a+1,a+cont+1,cmp);int ans=INF;for(int i=1;i<=cont;i++){if(a[i].flag==0)ans=min(ans,dp[m-a[i].mo]+a[i].time);elsedp[a[i].mo]=min(dp[a[i].mo],a[i].time);}if(ans!=INF)printf("%d\n",ans);elseprintf("oh no!\n");}
int main()
{while(~scanf("%d %d",&n,&m)){init();}return 0;
}

Coach's plan

时间限制: 2 Sec 内存限制: 128 MB
提交: 57 解决: 15
[提交][状态][讨论版]

题目描述

NEUACM team holds summer training every year. This year n students participate in the training, and we provide m computers. Everyone has a suitability (0<=suitability<=1000) to every computer.
Now our coach needs to draw up a plan, which satisfies:
1. every student uses one computer, and a computer can be used by no more than one student;
2. maximize the minimum suitability, which means if the answer is w, then everyone's suitability to his computer shouldn't be less than w.

Hint: huge input, please use fast IO.

输入

The first line contains n and m (1<=n<=m<=500), indicating the number of students and computers.

Next n lines describes every student's suitability to every computer, number in ith line and jth row means student[i]'s suitability to computer[j].

输出

A single number w in one line, which is described before.

样例输入

1 2

1 10

2 3

10 1 2

2 1 1

样例输出

找列最小水过的, 隔壁找行最小水过的, 然后重判后 game over 隔隔壁, 找行列最小依然坚挺 ; mmp

水过代码: 找行小列小, min 最小的,

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;int n,m,a[505][505],b[505],c[505];bool cmp(int x, int y){return x > y;
}int main(){while(~scanf("%d%d",&n,&m)){for(int i = 0; i < n; i ++){for(int j = 0; j < m; j ++)scanf("%d",&a[i][j]);}for(int i = 0; i < n; i ++){int mxx = a[i][0];for(int j = 1; j < m; j ++)mxx = max(a[i][j],mxx);b[i] = mxx;}for(int j = 0; j < m; j ++){int mxx = a[0][j];for(int i = 1; i < n; i ++)mxx = max(a[i][j],mxx);c[j] = mxx;}sort(b,b+n,cmp);sort(c,c+m,cmp);printf("%d\n",min(c[n-1],b[n-1]));}return 0;
}

正解是: 二分图匹配 + 二分查找;

把网格从一维变成二维匹配问题, 用到匈牙利算法,

比赛时, 考虑到二分图匹配, 没想到要用到二分查找,

要用链式前向星存储边l

1w 小心 , 一不小心就超时;

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};int n,m;
int maps[505][505];
struct Edge{int v,u;
}a[maxn];
int edge_cont;
int head[10005];
int match[10005];
bool used[10005];
void add(int u,int v)
{a[++edge_cont].v=v;a[edge_cont].u=head[u];head[u]=edge_cont;
}
int finds(int x)
{used[x]=1;for(int i=head[x];i!=-1;i=a[i].u)// 链式前向星 访问{int v=a[i].v;int w=match[v];if(w<0||!used[w]&&finds(w))// v 还没匹配 或者 能够找到更合适的{match[x]=v;match[v]=x;return 1;}}return 0;
}
int binary_match()//二分匹配
{int ans=0;mem(match,-1);for(int i=0;i<n;i++){if(match[i]<0)//未匹配{mem(used,0);if(finds(i))//可以找到合适的ans++;}}return ans;
}
bool judge(int x)
{edge_cont=0;mem(head,-1);for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(maps[i][j]>=x){add(i,j+n); //无向图add(j+n,i);}}}if(binary_match()==n)return true;return false;
}
void work()
{int ans;int l=0;int r=1001;while(r-l>1){int mid=(l+r)>>1;if(judge(mid))l=mid;elser=mid;}cout<<l<<endl;
}
int main()
{while(~scanf("%d %d",&n,&m)){mem(maps,0);mem(a,0);min_num=INF;max_num=0;for(int i=0;i<n;i++)for(int j=0;j<m;j++){scanf("%d",&maps[i][j]);}work();}return 0;
}

123