Codeforces Round #287 (Div. 2) E. Breaking Good [Dijkstra 最短路 优先队列]
2 seconds
256 megabytes
standard input
standard output
Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that is really difficult even for experienced gamers.
Walter William, the main character of the game, wants to join a gang called Los Hermanos (The Brothers). The gang controls the whole country which consists of n cities with m bidirectional roads connecting them. There is no road is connecting a city to itself and for any two cities there is at most one road between them. The country is connected, in the other words, it is possible to reach any city from any other city using the given roads.
The roads aren't all working. There are some roads which need some more work to be performed to be completely functioning.
The gang is going to rob a bank! The bank is located in city 1. As usual, the hardest part is to escape to their headquarters where the police can't get them. The gang's headquarters is in city n. To gain the gang's trust, Walter is in charge of this operation, so he came up with a smart plan.
First of all the path which they are going to use on their way back from city 1 to their headquarters n must be as short as possible, since it is important to finish operation as fast as possible.
Then, gang has to blow up all other roads in country that don't lay on this path, in order to prevent any police reinforcements. In case of non-working road, they don't have to blow up it as it is already malfunctional.
If the chosen path has some roads that doesn't work they'll have to repair those roads before the operation.
Walter discovered that there was a lot of paths that satisfied the condition of being shortest possible so he decided to choose among them a path that minimizes the total number of affected roads (both roads that have to be blown up and roads to be repaired).
Can you help Walter complete his task and gain the gang's trust?
The first line of input contains two integers n, m (2 ≤ n ≤ 105, 
), the number of cities and number of roads respectively.
In following m lines there are descriptions of roads. Each description consists of three integers x, y, z (1 ≤ x, y ≤ n, 
) meaning that there is a road connecting cities number x and y. If z = 1, this road is working, otherwise it is not.
In the first line output one integer k, the minimum possible number of roads affected by gang.
In the following k lines output three integers describing roads that should be affected. Each line should contain three integers x, y, z (1 ≤ x, y ≤ n, ), cities connected by a road and the new state of a road. z = 1 indicates that the road between cities x and y should be repaired and z = 0 means that road should be blown up.
You may output roads in any order. Each affected road should appear exactly once. You may output cities connected by a single road in any order. If you output a road, it's original state should be different from z.
After performing all operations accroding to your plan, there should remain working only roads lying on some certain shortest past between city 1 and n.
If there are multiple optimal answers output any.
2 1
1 2 0
1
1 2 1
4 4
1 2 1
1 3 0
2 3 1
3 4 1
3
1 2 0
1 3 1
2 3 0
8 9
1 2 0
8 3 0
2 3 1
1 4 1
8 7 0
1 5 1
4 6 1
5 7 0
6 8 0
3
2 3 0
1 5 0
6 8 1
In the first test the only path is 1 - 2
In the second test the only shortest path is 1 - 3 - 4
In the third test there are multiple shortest paths but the optimal is 1 - 4 - 6 - 8
有n个城市，m条边。让你求出来1到n的最短路，但是要求，保证最短路的时候，要求权值最大。
最开始没反应过来是求最短路问题，直接bfs，，，跪惨。。。。
| 9557892 | 2015-01-26 11:37:48 | njczy2010 | E - Breaking Good | GNU C++ | Accepted | 124 ms | 19100 KB |
| 9557655 | 2015-01-26 11:12:44 | njczy2010 | E - Breaking Good | GNU C++ | Accepted | 623 ms | 19300 KB |
| 9557440 | 2015-01-26 10:51:36 | njczy2010 | E - Breaking Good | GNU C++ | Wrong answer on test 8 | 31 ms | 6100 KB |
| 9557233 | 2015-01-26 10:29:53 | njczy2010 | E - Breaking Good | GNU C++ | Wrong answer on test 8 | 15 ms | 6200 KB |
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<string> #define N 100005
#define M 105
#define mod 1000000007
//#define p 10000007
#define mod2 1000000000
#define ll long long
#define LL long long
#define eps 1e-6
#define inf 100000000
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b) using namespace std; int n,m;
int x[N],y[N],z[N];
int tot; typedef struct
{
int a,b;
int f;
}AA; AA ans[N]; typedef struct
{
int to;
int f;
int indexb;
}PP; vector<PP>bian[N];
int vis[N];
int Step[N];
int Cnt[N]; struct QQ
{
friend bool operator < (QQ n1,QQ n2)
{
if(n1.step==n2.step){
return n1.cnt>n2.cnt;
}
else{
return n1.step>n2.step;
}
}
int index;
int cnt;
int indexn;
int pre;
int step;
}; void out(QQ te); void ini()
{
int i;
tot=;
memset(vis,,sizeof(vis));
PP te;
for(i=;i<=n;i++){
bian[i].clear();
Step[i]=inf;
Cnt[i]=inf;
}
Step[]=;
Cnt[]=;
for(i=;i<=m;i++){
scanf("%d%d%d",&x[i],&y[i],&z[i]);
te.to=y[i];te.f=z[i];te.indexb=i;
bian[ x[i] ].push_back(te);
te.to=x[i];
bian[ y[i] ].push_back(te);
}
} typedef struct
{
int in;
int pre;
}NN; NN node[N*];
int totnode; void bfs()
{
priority_queue <QQ>q;
QQ te,nt;
te.index=;
te.cnt=;
te.step=;
totnode=;
te.indexn=-;
te.pre=-;
q.push(te);
int st;
PP yy;
vector<PP>::iterator it;
while(q.size()>=)
{
te=q.top();
q.pop();
if(te.index==n){
out(te);
return;
}
if(vis[te.index]==) continue;
vis[te.index]=;
st=te.index;
for(it=bian[st].begin();it!=bian[st].end();it++)
{
yy=*it;
if(vis[yy.to]==) continue;
nt.index=yy.to;
nt.cnt=te.cnt;
nt.pre=te.index;
nt.step=te.step+;
if(yy.f==){
nt.cnt++;
}
if(nt.step>Step[nt.index]) continue;
if(nt.step==Step[nt.index] && nt.cnt>Cnt[nt.index]) continue;
Step[nt.index]=nt.step;
Cnt[nt.index]=nt.cnt;
nt.indexn=totnode;
node[totnode].in=yy.indexb;
node[totnode].pre=te.indexn;
totnode++;
q.push(nt);
}
}
} void solve()
{
bfs();
} void out(QQ te)
{
int i;
int vis2[N];
memset(vis2,,sizeof(vis2));
i=te.indexn;
int temp;
while(i!=-)
{
temp=node[i].in;
vis2[temp]=;
i=node[i].pre;
}
for(i=;i<=m;i++){
if(vis2[i]==){
if(z[i]==){
ans[tot].a=x[i];
ans[tot].b=y[i];
ans[tot].f=;
tot++;
}
}
else{
if(z[i]==){
ans[tot].a=x[i];
ans[tot].b=y[i];
ans[tot].f=;
tot++;
}
}
}
printf("%d\n",tot);
for(i=;i<tot;i++){
printf("%d %d %d\n",ans[i].a,ans[i].b,ans[i].f);
}
} int main()
{
//freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
//scanf("%d",&T);
//for(int ccnt=1;ccnt<=T;ccnt++)
//while(T--)
while(scanf("%d%d",&n,&m)!=EOF)
{
ini();
solve();
// out();
}
return ;
}
一开始wa在test8的代码：
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<string> #define N 100005
#define M 105
#define mod 1000000007
//#define p 10000007
#define mod2 1000000000
#define ll long long
#define LL long long
#define eps 1e-6
#define inf 100000000
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b) using namespace std; int n,m;
int x[N],y[N],z[N];
int tot; typedef struct
{
int a,b;
int f;
}AA; AA ans[N]; typedef struct
{
int to;
int f;
int indexb;
}PP; vector<PP>bian[N];
int vis[N];
int Step[N];
int Cnt[N]; struct QQ
{
friend bool operator < (QQ n1,QQ n2)
{
if(n1.step==n2.step){
return n1.cnt>n2.cnt;
}
else{
return n1.step>n2.step;
}
}
int index;
int cnt;
string s;
int indexn;
int pre;
int step;
}; void out(QQ te); void ini()
{
int i;
tot=;
memset(vis,,sizeof(vis));
PP te;
for(i=;i<=n;i++){
bian[i].clear();
Step[i]=inf;
Cnt[i]=inf;
}
Step[]=;
Cnt[]=;
for(i=;i<=m;i++){
scanf("%d%d%d",&x[i],&y[i],&z[i]);
te.to=y[i];te.f=z[i];te.indexb=i;
bian[ x[i] ].push_back(te);
te.to=x[i];
bian[ y[i] ].push_back(te);
}
} typedef struct
{
int in;
int pre;
}NN; NN node[N*];
int totnode; void bfs()
{
priority_queue <QQ>q;
QQ te,nt;
te.index=;
te.cnt=;
te.step=;
te.s="";
totnode=;
te.indexn=-;
te.pre=-;
//vis[1]=1;
q.push(te);
int st;
PP yy;
char ss;
vector<PP>::iterator it;
while(q.size()>=)
{
te=q.top();
// printf(" index=%d cnt=%d",te.index,te.cnt);
// cout<<" s="<<te.s<<endl;
q.pop();
if(te.index==n){
out(te);
return;
}
if(vis[te.index]==) continue;
vis[te.index]=;
st=te.index;
for(it=bian[st].begin();it!=bian[st].end();it++)
{
yy=*it;
if(vis[yy.to]==) continue;
nt.index=yy.to;
ss=yy.indexb+'';
nt.s=te.s+ss;
// vis[yy.indexb]=1;
//vis[nt.index]=1;
nt.cnt=te.cnt;
nt.pre=te.index;
nt.step=te.step+;
if(yy.f==){
nt.cnt++;
}
if(nt.step>Step[nt.index]) continue;
if(nt.step==Step[nt.index] && nt.cnt>Cnt[nt.index]) continue;
Step[nt.index]=nt.step;
Cnt[nt.index]=nt.cnt;
nt.indexn=totnode;
node[totnode].in=yy.indexb;
node[totnode].pre=te.indexn;
totnode++;
q.push(nt); // printf(" index=%d cnt=%d",nt.index,nt.cnt);
// cout<<" s="<<nt.s<<endl;
}
}
} void solve()
{
bfs();
} void out(QQ te)
{
int l=te.s.length();
int i;
int vis2[N];
memset(vis2,,sizeof(vis2));
// cout<<"te.s="<<te.s<<" l="<<l<<endl;
i=te.indexn;
int temp;
while(i!=-)
{
temp=node[i].in;
vis2[temp]=;
i=node[i].pre;
}
// for(i=0;i<l;i++){
// vis2[ te.s[i]-'0' ]=1;
// }
for(i=;i<=m;i++){
if(vis2[i]==){
if(z[i]==){
ans[tot].a=x[i];
ans[tot].b=y[i];
ans[tot].f=;
tot++;
}
}
else{
if(z[i]==){
ans[tot].a=x[i];
ans[tot].b=y[i];
ans[tot].f=;
tot++;
}
}
}
printf("%d\n",tot);
for(i=;i<tot;i++){
printf("%d %d %d\n",ans[i].a,ans[i].b,ans[i].f);
}
} int main()
{
//freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
//scanf("%d",&T);
//for(int ccnt=1;ccnt<=T;ccnt++)
//while(T--)
while(scanf("%d%d",&n,&m)!=EOF)
{
ini();
solve();
// out();
}
return ;
}
Codeforces Round #287 (Div. 2) E. Breaking Good [Dijkstra 最短路 优先队列]的更多相关文章
- Codeforces Round #287 (Div. 2) E. Breaking Good 最短路
题目链接: http://codeforces.com/problemset/problem/507/E E. Breaking Good time limit per test2 secondsme ...
- Codeforces Round #287 (Div. 2) E. Breaking Good 路径记录！！！+最短路+堆优化
E. Breaking Good time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...
- 贪心 Codeforces Round #287 (Div. 2) A. Amr and Music
题目传送门 /* 贪心水题 */ #include <cstdio> #include <algorithm> #include <iostream> #inclu ...
- Codeforces Round #287 (Div. 2) C. Guess Your Way Out! 思路
C. Guess Your Way Out! time limit per test 1 second memory limit per test 256 megabytes input standa ...
- CodeForces Round #287 Div.2
A. Amr and Music (贪心) 水题,没能秒切,略尴尬. #include <cstdio> #include <algorithm> using namespac ...
- Codeforces Round #287 (Div. 2) C. Guess Your Way Out! 水题
C. Guess Your Way Out! time limit per test 1 second memory limit per test 256 megabytes input standa ...
- Codeforces Round #287 (Div. 2) B. Amr and Pins 水题
B. Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- Codeforces Round #287 (Div. 2) A. Amr and Music 水题
A. Amr and Music time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Codeforces Round #287 (Div. 2) D. The Maths Lecture [数位dp]
传送门 D. The Maths Lecture time limit per test 1 second memory limit per test 256 megabytes input stan ...
随机推荐
- CAD命令标志
CAD命令标志 主标识:(常用的)ACRX_CMD_MODAL 在别的命令执行的时候该命令不会在其中执行.ACRX_CMD_TRANSPARENT 命令可以再其它命令中执行,但在该标志下ads_sss ...
- chrom控制台常用方法
console.assert对输入的表达式进行断言,只有表达式为false时,才输出相应的信息到控制台 . console.count(这个方法非常实用哦)当你想统计代码被执行的次数 console. ...
- AEE加密解密
from Crypto.Cipher import AESfrom binascii import b2a_hex, a2b_hex class AesHandler(object): def ...
- 【转】IntelliJ 创建main函数快捷
http://blog.csdn.net/tiantiandjava/article/details/42269173 今天偶然发现了IntelliJ中 创建main函数的快捷键,依次还有for循环, ...
- shell脚本，计算1+2+3+....100等于多少？
第一种方法,通过for循环来计算[root@localhost wyb]# cat yibai.sh #!/bin/bash #从1+++...100的结果 i= ` do sum=$(($sum+i ...
- UVa-213-信息解码
这题的话,我们只要理解题意,应该就不算很难. 我们可以开一个二维数组,用来存放对应的编码字符,第一个下表是length,第二个下标是value,这样一来,我们在读入数据的时候就进行处理,然后想要使用的 ...
- （22）zabbix触发器依赖关系详解
概述 zabbix触发器可以设置依赖性,例如我配置了两个触发器,一个触发器定义www.ttlsa.com这个HOST是否在运行中,另一个是www.ttlsa.com的网络是否通畅. 假如网络出现故障, ...
- linux-MySQL基本指令-增删改查
常用指令 指令作用 指令 查看/查询 show,select,desc 创建 create 删除 drop,delete,truncate 切换/进入 use 添加记录 insert 查询类 查看数据 ...
- link与@import导入css样式区别
XML/HTML代码<link rel="stylesheet" rev="stylesheet" href="CSS文件" type ...
- perl学习之六：变量作用域
变量作用域 (一)包变量 $x = 1 这里,$x是个包变量.关于包变量,有2件重要的事情要了解: 1)假如没有其他申明,变量就是包变量:2)包变量总是全局的. 全局意味着包变量在每个程序里总可访问到 ...