HDU--3605 Escape (多重匹配)
来源:互联网 发布:网络歌曲改编歌词 编辑:程序博客网 时间:2024/06/05 02:02
Problem Description
2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.
Input
More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000
Output
Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.
If you can output YES, otherwise output NO.
Sample Input
1 1112 21 01 01 1
Sample Output
YESNO
思路:用一个二维数组存与星球匹配的人员的编号,若有人无法匹配,则输出NO
AC代码:
#include<cstdio>#include<cstring>const int N=100010;bool vis[15];int map[N][15],mat[15][N];int cnt[15],vol[15];int n,m;bool find(int x){ int i,j; for(i=0;i<m;i++) { if(!vis[i]&&map[x][i]) { vis[i]=true; if(cnt[i]<vol[i]) { mat[i][cnt[i]++]=x; return true; } for(j=0;j<cnt[i];j++) { if(find(mat[i][j])) { mat[i][j]=x; return true; } } } } return false;}int MMG(){ int i; memset(cnt,0,sizeof(cnt)); for(i=0;i<n;i++) { memset(vis,false,sizeof(vis)); if(!find(i)) return false; } return true;}int main(){ int i,j; while(scanf("%d%d",&n,&m)!=EOF) { for(i=0;i<n;i++) { for(j=0;j<m;j++) scanf("%d",&map[i][j]); } for(i=0;i<m;i++) scanf("%d",&vol[i]); if(MMG()) printf("YES\n"); else printf("NO\n"); } return 0;}
0 0
- HDU--3605 Escape (多重匹配)
- hdu 3605 Escape(多重匹配)
- hdu 3605 Escape (二分图多重匹配)
- hdu 3605 Escape(多重匹配)
- hdu 3605 Escape【二分图多重匹配】
- HDU 3605 Escape【二分图多重匹配】
- hdu 3605 Escape(二分图多重匹配问题)
- HDU 3605 Escape (二分图的多重匹配)
- hdu 3605 Escape (二分图的多重匹配)
- hdu 3605 Escape 多重匹配模板 (Hungary ) | 最大流
- hdu 3605 Escape (二分图多重匹配)
- HDU 3605 Escape (二分图多重匹配模板)
- HDU 3605 Escape (二分图多重匹配模板)
- HDU 3605 Escape(多重匹配之多对多的匹配)
- 二分图匹配 ( 多重匹配&&Hungarian)——Escape ( HDU 3605 )
- HDOJ 3605 - Escape 二分图多重匹配
- hdu3605 Escape (多重匹配)
- HDU3605 Escape 多重匹配
- 广东工业大学 数据结构实验报告 题目: 稀疏矩阵
- 关键在于第四个对象
- hdu 2043 密码 20140728.cpp
- 各种排序的实现:希尔 归并 堆排 快排
- Dom编程(Window 对象)
- HDU--3605 Escape (多重匹配)
- poj 2724 Purifying Machine 二分图最大匹配
- IOS调用系统震动和系统声音
- HBase Capacity Planning
- 大数据公司实践零售O2O:打通线上线下,全触点大会员
- POJ 1751 Prim
- HDU 3657 Game | 最小割 (补)
- EXT4.1 tabpanel
- Irrlicht例程05:创造一个有声世界(下)