标签：os io for art ar 时间 amp size
用floyd超时了。。。注定的事情。。。题意：看案例就跑出来了。。不需要看题了把。。#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#include<vector>
const int INF =1999299;
int minn(int a,int b)
{
return a>b?b:a;
}
#define N 510
#define M 251000
int n, m;//n m 为点数和边数
struct Edge
{
int from, to, val,nex;
bool sign;//是否为桥
} edge[M<<1];
int head[N], edgenum;
void add(int u, int v,int val) //边的起点和终点
{
Edge E= {u, v, val,head[u], false};
edge[edgenum] = E;
head[u] = edgenum++;
}
int DFN[N], Low[N], Stack[N], top, Time; //Low[u]是点集{u点及以u点为根的子树} 中(所有反向弧)能指向的(离根最近的祖先v) 的DFN[v]值(即v点时间戳)
int taj;//连通分支标号，从1开始
int Belong[N];//Belong[i] 表示i点属于的连通分支
bool Instack[N];
vector<int> bcc[N]; //标号从1开始
void tarjan(int u ,int fa)
{
DFN[u] = Low[u] = ++ Time ;
Stack[top ++ ] = u ;
Instack[u] = 1 ;
for (int i = head[u] ; ~i ; i = edge[i].nex )
{
int v = edge[i].to;
if(DFN[v] == -1)
{
tarjan(v , u);
Low[u] = minn(Low[u] ,Low[v]) ;
if(DFN[u] < Low[v])
{
edge[i].sign = 1;//为割桥
}
}
else if(Instack[v])
{
Low[u] = minn(Low[u] ,DFN[v]) ;
}
}
if(Low[u] == DFN[u])
{
int now;
taj ++ ;
bcc[taj].clear();
do
{
now = Stack[-- top] ;
Instack[now] = 0 ;
Belong [now] = taj ;
bcc[taj].push_back(now);
}
while(now != u) ;
}
}
void tarjan_init(int all)
{
memset(DFN, -1, sizeof(DFN));
memset(Instack, 0, sizeof(Instack));
top = Time = taj = 0;
for(int i=1; i<=all; i++)
if(DFN[i]==-1 )
tarjan(i, i); //注意开始点标！！！
}
vector<int>G[N];
int du[N];
void suodian()
{
memset(du, 0, sizeof(du));
for(int i = 1; i <= taj; i++)
G[i].clear();
for(int i = 0; i < edgenum; i++)
{
int u = Belong[edge[i].from], v = Belong[edge[i].to];
if(u!=v)
{
G[u].push_back(v), du[v]++;
}
}
}
int d[N][N],cost[N][N];
void Dijkstra(int n,int beg)
{
bool vis[N];
for(int i=0; i<=n; i++)
{
d[beg][i]=INF;
vis[i]=false;
}
d[beg][beg]=0;
for(int j=1; j<=n; j++)
{
int k=-1;
int Min=INF;
for(int i=0; i<=n; i++)
if(!vis[i]&&d[beg][i]<Min)
{
Min=d[beg][i];
k=i;
}
if(k==-1)break;
vis[k]=true;
for(int i=0; i<=n; i++)
if(!vis[i]&&d[beg][k]+cost[k][i]<d[beg][i])
{
d[beg][i]=d[beg][k]+cost[k][i];
}
}
}
void init()
{
memset(head, -1, sizeof(head));
edgenum=0;
}
int main()
{
// freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&m),n+m)
{
init();
int a,b,val;
for(int i=0; i<=n; i++)
{
for(int j=0; j<=n; j++)
{
cost[i][j]=INF;
}
}
for(int i=0; i<m; i++)
{
scanf("%d%d%d",&a,&b,&val);
add(a,b,val);
}
tarjan_init(n);
suodian();//缩点
int q;
for(int i=0; i<m; i++)//重新建图
{
int u = Belong[edge[i].from], v = Belong[edge[i].to];
if(u==v)
continue;
cost[u][v]=minn(cost[u][v],edge[i].val);
}
scanf("%d",&q);
int use[N];
memset(use,0,sizeof(use));
for(int i=0; i<q; i++)
{
int a,b;
scanf("%d%d",&a,&b);
int u = Belong[a], v = Belong[b];
if(!use[u])
{
Dijkstra(taj,u);
use[u]=1;
}
if(u==v)
printf("0\n");
else if(d[u][v]==INF)
printf("Nao e possivel entregar a carta\n");
else
printf("%d\n",d[u][v]);
}
printf("\n");
}
return 0;
}POJ 3114 Countries in War 强连通+最短路,布布扣,bubuko.com
POJ 3114 Countries in War 强连通+最短路
标签：os io for art ar 时间 amp size
原文地址：http://blog.csdn.net/kewowlo/article/details/38351085
