Codeforces Round #333 (Div. 2) B. Approximating a Constant Range
2016-04-06 22:02
239 查看
C. The Two Routes
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
In Absurdistan, there are n towns (numbered 1 through n)
and m bidirectional railways. There is also an absurdly simple road network — for each pair of different towns x and y,
there is a bidirectional road between towns x and y if
and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town n,
and don't make any stops on the way (but they can wait in town n). The train can move only along railways and the bus can move only
along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the
same town (except town n) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town n (the maximum of arrival times
of the bus and the train)? Note, that bus and train are not required to arrive to the town n at the same moment of time, but are allowed
to do so.
Input
The first line of the input contains two integers n and m (2 ≤ n ≤ 400, 0 ≤ m ≤ n(n - 1) / 2) —
the number of towns and the number of railways respectively.
Each of the next m lines contains two integers u and v,
denoting a railway between towns u and v (1 ≤ u, v ≤ n, u ≠ v).
You may assume that there is at most one railway connecting any two towns.
Output
Output one integer — the smallest possible time of the later vehicle's arrival in town n. If it's impossible for at least one of the
vehicles to reach town n, output - 1.
Examples
input
output
input
output
input
output
Note
In the first sample, the train can take the route
![](http://codeforces.com/predownloaded/45/30/45307ea441d2996b4e3db04de8169f6600202297.png)
and
the bus can take the route
![](http://codeforces.com/predownloaded/17/34/17348a86f2090cbcfe747091924a85e6176e16c0.png)
.
Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4.
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
In Absurdistan, there are n towns (numbered 1 through n)
and m bidirectional railways. There is also an absurdly simple road network — for each pair of different towns x and y,
there is a bidirectional road between towns x and y if
and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town n,
and don't make any stops on the way (but they can wait in town n). The train can move only along railways and the bus can move only
along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the
same town (except town n) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town n (the maximum of arrival times
of the bus and the train)? Note, that bus and train are not required to arrive to the town n at the same moment of time, but are allowed
to do so.
Input
The first line of the input contains two integers n and m (2 ≤ n ≤ 400, 0 ≤ m ≤ n(n - 1) / 2) —
the number of towns and the number of railways respectively.
Each of the next m lines contains two integers u and v,
denoting a railway between towns u and v (1 ≤ u, v ≤ n, u ≠ v).
You may assume that there is at most one railway connecting any two towns.
Output
Output one integer — the smallest possible time of the later vehicle's arrival in town n. If it's impossible for at least one of the
vehicles to reach town n, output - 1.
Examples
input
4 2 1 3 3 4
output
2
input
4 6
1 21 31 4
2 32 4
3 4
output
-1
input
5 5
4 23 5
4 5
5 1
1 2
output
3
Note
In the first sample, the train can take the route
![](http://codeforces.com/predownloaded/45/30/45307ea441d2996b4e3db04de8169f6600202297.png)
and
the bus can take the route
![](http://codeforces.com/predownloaded/17/34/17348a86f2090cbcfe747091924a85e6176e16c0.png)
.
Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4.
#include <stdio.h> #include <string.h> #include <algorithm> #define INF 1 << 29 using namespace std; int map1[410][410],map2[410][410]; //记录火车的路和大巴的路 int n,m; void dijkstra(int map[][410]) { int res[410]; //记录结果 int vis[410]; //访问标记 int v,min; //每相连点的最小下标和最短的路 for(int i = 1 ; i <= n ; i++ ) //初始化 { vis[i] = 0; res[i] = map[1][i]; } for (int i = 1 ; i <= n ; i++ ) { min = 1<< 29; for (int j = 1 ; j <= n ; j++ ) { if (vis[j] == 0 && res[j] < min ) //遍历,找出与这个点相连的最小路 { v = j; min = res[j]; } } vis[v] = 1; for (int j = 1 ; j <= n ; j++ ) { if(vis[j] == 0 && res[j] > map[v][j] + res[v]) //判断 { res[j] = map[v][j] +res[v]; } } } if (res == INF) //如果最后的值没有改变 说明不可能 { printf("-1\n"); return ; } else printf("%d\n",res ); } int main() { while (~scanf("%d%d",&n,&m ) ) { for (int i = 1 ; i <= n ; i++ ) //初始化 { for (int j = 1 ; j <= n ; j++ ) { if ( i == j) { map1[i][j] = 0; map1[i][j] = 0; } else { map1[i][j] = map1[j][i] = INF; map2[i][j] = map2[j][i] = INF; } } } int flag = 0; for (int i = 0 ; i < m ; i++ ) { int a,b; scanf("%d%d",&a,&b); if ( (a == 1 && b == n) || (a == n && b == 1) ) //必有一条路可以直达,所以判断是火车还是巴士 { flag = 1; } map1[a][b] = 1; map1[b][a] = 1; } for (int i = 1 ; i <= n ; i ++ ) { for (int j = 1 ; j <= n ; j++ ) { if ( map1[i][j] == INF || map1[i][j] == 0) { map2[i][j] = 1; map2[j][i] = 1; } } } if (m == n*(n-1)/2) //如果只有火车有路径 { printf("-1\n"); continue; } if (flag == 1) //如果火车可以直达 { dijkstra(map2); } else if (flag == 0) //如果大巴可以直达 { dijkstra(map1); } } return 0; }
相关文章推荐
- Android开发之发送邮件功能的实现(源代码分享)
- Objective-C 打印九九乘法表
- DTAM: Dense Tracking and Mapping(一)
- android之LayoutInflater
- MSB3073 命令“platforms\android\cordova\clean.bat”已退出,代码为 2。
- iOS_主要的函数
- Log4j2 JDBCAppender的使用
- iOS设计中不同屏幕适配的方法-登陆界面
- Qt for Android开发中消除NDK的黑屏启动页面
- iOS SDWEBImage和collectionView的组合,以及collectionView的随意间距设置
- android大文件上传
- On Memory Leaks in Java and in Android.
- 终于懂了:Delphi消息的Result域出现的原因——要代替回调函数的返回值!(MakeObjectInstance不会帮助处理(接收)消息回调函数的返回值)
- Android 引导使用说明、标签功能的封装库
- Android 操作数据库的框架——greenDAO的学习
- 闲说App的接口测试
- Android学习之CardView文档翻译
- Android的桌面小图标Widget的学习
- iOS中WebSocket的使用
- 开源图片加载库universal-image-loader使用