【POJ 1125】Stockbroker Grapevine
id=1125">【POJ 1125】Stockbroker Grapevine
最短路 只是这题数据非常水。
。
主要想大牛们试试南阳OJ同题 链接例如以下:
http://acm.nyist.net/JudgeOnline/talking.php?pid=426&page=2
数据增大非常多 用到非常多东西才干过 (弱没过,。。
这题就是求最短路寻找全部通路中最大权的最小值外加考验英语水平……
Floyd 208K 0MS 1162B
#include
using namespace std;
int dis[111][111],n;
void Floyd()
{
int i,j,k,tmax,mmax,f;
for(k = 1; k <= n; ++k)
for(i = 1; i <= n; ++i)
for(j = 1; j <= n; ++j)
if(dis[i][j] > dis[i][k] + dis[k][j])
dis[i][j] = dis[i][k] + dis[k][j];
mmax = INF;
for(i = 1; i <= n; ++i)
{
f = 1;
tmax = 0;
for(j = 1; j <= n; ++j)
{
if(i == j) continue;
if(dis[i][j] == INF) f = 0;
tmax = max(tmax,dis[i][j]);
}
if(f && tmax < mmax)
{
k = i;
mmax = tmax;
}
}
if(mmax != INF) printf("%d %d\n",k,mmax);
else puts("disjoint");
}
int main()
{
int i,k,v;
while(~scanf("%d",&n) && n)
{
memset(dis,INF,sizeof(dis));
for(i = 1; i <= n; ++i)
{
scanf("%d",&k);
while(k--)
{
scanf("%d",&v);
scanf("%d",&dis[i][v]);
}
}
Floyd();
}
return 0;
}
Dijkstra 168K 0MS 1491B
#include
using namespace std;
typedef struct Edge
{
int v,w,next;
}Edge;
Edge eg[11111];
int head[111],dis[111],n,tp;
bool vis[111];
int Dijkstra(int u)
{
memset(dis,INF,sizeof(dis));
memset(vis,0,sizeof(vis));
dis[u] = 0;
int m,p,i,j;
for(i = 1; i <= n; ++i)
{
p = -1;
m = INF;
for(j = 1; j <= n; ++j)
{
if(!vis[j] && dis[j] < m)
{
p = j;
m = dis[j];
}
}
if(i == n || p == -1) break;
vis[p] = 1;
for(j = head[p]; j != -1; j = eg[j].next)
{
if(!vis[eg[j].v] && dis[eg[j].v] > dis[p] + eg[j].w)
dis[eg[j].v] = dis[p] + eg[j].w;
}
}
if(p == -1) return INF;
return dis[p];
}
int main()
{
int i,k,m,t;
while(~scanf("%d",&n) && n)
{
m = INF;
tp = 0;
memset(head,-1,sizeof(head));
for(i = 1; i <= n; ++i)
{
scanf("%d",&k);
while(k--)
{
scanf("%d %d",&eg[tp].v,&eg[tp].w);
eg[tp].next = head[i];
head[i] = tp++;
}
}
k = 0;
for(i = 1; i <= n; ++i)
{
t = Dijkstra(i);
if(t < m)
{
k = i;
m = t;
}
}
if(k)
printf("%d %d\n",k,m);
else puts("disjoint");
}
return 0;
}
SPFA 180K 0MS 1668B
#include
using namespace std;
typedef struct Edge
{
int v,w,next;
}Edge;
Edge eg[11111];
int head[111],dis[111],n,tp;
bool vis[111];
int SPFA(int u)
{
memset(dis,INF,sizeof(dis));
memset(vis,0,sizeof(vis));
dis[u] = 0;
queue <int> q;
q.push(u);
int v,w,i,p,m;
while(!q.empty())
{
p = q.front();
q.pop();
vis[u] = 0;
for(i = head[p]; i != -1; i = eg[i].next)
{
v = eg[i].v;
w = eg[i].w;
if(dis[v] > dis[p] + w)
{
dis[v] = dis[p] + w;
if(!vis[v])
{
vis[v] = 1;
q.push(v);
}
}
}
}
m = 0;
for(i = 1; i <= n; ++i)
{
if(i == u) continue;
if(dis[i] == INF) return INF;
m = max(m,dis[i]);
}
return m;
}
int main()
{
int i,k,m,t;
while(~scanf("%d",&n) && n)
{
m = INF;
tp = 0;
memset(head,-1,sizeof(head));
for(i = 1; i <= n; ++i)
{
scanf("%d",&k);
while(k--)
{
scanf("%d %d",&eg[tp].v,&eg[tp].w);
eg[tp].next = head[i];
head[i] = tp++;
}
}
k = 0;
for(i = 1; i <= n; ++i)
{
t = SPFA(i);
if(t < m)
{
k = i;
m = t;
}
}
if(k)
printf("%d %d\n",k,m);
else puts("disjoint");
}
return 0;
}
【POJ 1125】Stockbroker Grapevine的更多相关文章
- POJ 1125:Stockbroker Grapevine
Stockbroker Grapevine Time Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %I64d & %I64 ...
- bzoj 2295: 【POJ Challenge】我爱你啊
2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec Memory Limit: 128 MB Description ftiasch是个十分受女生欢迎的同学,所以 ...
- 【链表】BZOJ 2288: 【POJ Challenge】生日礼物
2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 382 Solved: 111[Submit][S ...
- BZOJ2288: 【POJ Challenge】生日礼物
2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 284 Solved: 82[Submit][St ...
- BZOJ2293: 【POJ Challenge】吉他英雄
2293: [POJ Challenge]吉他英雄 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 80 Solved: 59[Submit][Stat ...
- BZOJ2287: 【POJ Challenge】消失之物
2287: [POJ Challenge]消失之物 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 254 Solved: 140[Submit][S ...
- BZOJ2295: 【POJ Challenge】我爱你啊
2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 126 Solved: 90[Submit][Sta ...
- BZOJ2296: 【POJ Challenge】随机种子
2296: [POJ Challenge]随机种子 Time Limit: 1 Sec Memory Limit: 128 MBSec Special JudgeSubmit: 114 Solv ...
- BZOJ2292: 【POJ Challenge 】永远挑战
2292: [POJ Challenge ]永远挑战 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 513 Solved: 201[Submit][ ...
随机推荐
- redhat 6.3 64位安装中文输入法全过程记录
首先,修改/etc/profile文件,在末尾增加两行: export LC_ALL="zh_CN.UTF-8" export LANG="zh_CN.UTF-8&quo ...
- xcode 5与ios 7的屏幕适配问题
#define DEVICE_IS_IPHONE5 ([[UIScreen mainScreen] bounds].size.height == 568) float height = DEVICE_ ...
- wordpress参考网站
wordpress大学http://www.wpdaxue.com/post-tags-and-categories-for-pages.html
- TCP 协议三次握手过程解析带实例
TCP(Transmission Control Protocol) 传输控制协议 TCP是主机对主机层的传输控制协议,提供可靠的连接服务,采用三次握手确认建立一个连接: 位码即tcp标志位,有6种标 ...
- 委托-异步调用-泛型委托-匿名方法-Lambda表达式-事件【转】
1. 委托 From: http://www.cnblogs.com/daxnet/archive/2008/11/08/1687014.html 类是对象的抽象,而委托则可以看成是函数的抽象.一个委 ...
- dojo demo, server验证username是否已经被使用
这个demo有助于理解JS与server的协同工作. 文档结构如上图. 主要是三个文件: main.js table.html validateUserName.jsp (代码见文章末尾) 页面打 ...
- IDA strings view 中文字符的显示
具体原理不清楚,在网上找了找,记录下. 第一步:将ida.cfg中cpp866 version的AsciiStringChars注释掉,把full version的AsciiStringChars取消 ...
- 子级Repeater获取 父级Repeater
第一种方法,子级Repeater中绑定父级的某个字段: <%# DataBinder.Eval((Container.NamingContainer.NamingContainer as Rep ...
- web Form 表单method="get" method="post" 区别
get和post方法的不同 在B/S应用程序中,前台与后台的数据交互,都是通过HTML中Form表单完成的.Form提供了两种数据传输的方式——get和post.虽然它们都是数据的提交方式,但是在实际 ...
- Oracle触发器trigger3利用时间限制用户输入
--触发器的应用限制用户写入 --具体功能:在写入一个表之前,限制必须要在周一到周5和工作时间8:00~18:00 create or replace trigger tri3 before inse ...