poj 3268 Silver Cow Party

/* Name: poj 3268 Silver Cow Party Author: UnimenSun Date: 24/05/2011 23:08 Description: 变形的最短路径*//*解题报告：变形的最短路径分析：以样例输入为例：1、先求出各点到2点的最短路径放到数组dis2中2、再求出2点到各点的最短路径放到数组dis1中3、求出max(dis1 + dis2)，即为所求关键是第1步怎样求各点到2点的最短路径，用Floyd由于数据量太大会超时，用SPFA、Dijkistra会多次反复的用该算法也会造成超时，其实只到图中的有向边反转，反转后求2点到各点的最短距离，就相当于求各点到2点的距离*///代码一：SPFA算法#include <iostream>#include <queue>using namespace std;const int MAXN = 2100000;int g[1010][1010];int n, m, x;int dis1[1010], dis2[1010];void spfa(int *dis){int i;for(i=1; i<=n; ++i)dis[i] = MAXN;queue<int> que;que.push(x);dis[x] = 0;while(!que.empty()){int x = que.front();que.pop();for(i=1; i<=n; ++i){if(g[x][i]!=MAXN && dis[i]>dis[x]+g[x][i]){dis[i] = dis[x] + g[x][i];que.push(i);}}}}//有向边转置，其实就是矩阵转置void Traverse(){int i, j;for(i=1; i<=n; ++i){for(j=1; j<=i; ++j){int temp = g[i][j];g[i][j] = g[j][i];g[j][i] = temp;}}}int main(){int i, j;while(cin>>n>>m>>x){//读数据并建图for(i=1; i<=n; ++i)for(j=1; j<=n; ++j)if(i == j)g[i][j] = 0;elseg[i][j] = MAXN;while(m--){int a, b, t;cin>>a>>b>>t;g[a][b] = t;}spfa(dis1);Traverse();spfa(dis2);int nMax = 0;for(i=1; i<=n; ++i){if(i != x){if(nMax<dis1[i]+dis2[i] && dis1[i]+dis2[i]<MAXN)nMax = dis1[i] + dis2[i];}}cout<<nMax<<endl;}return 0;}//代码二：Dijkistra求单源最短路径#include <iostream>#include <cstring>using namespace std;const int MAXN = 2100000;int g[1010][1010];int n, m, x;int dis1[1010], dis2[1010];void Dijkistra(int *dis){int i, j;int visited[1010];memset(visited, 0, sizeof(visited));for(i=1; i<=n; ++i){dis[i] = g[x][i];visited[i] = 0;}dis[x] = 0;visited[x] = 1;for(i=1; i<n; ++i){int index=0;int nMin = MAXN;for(j=1; j<=n; ++j){if(!visited[j] && nMin>dis[j]){index = j;nMin = dis[j];}}visited[index] = 1;for(j=1; j<=n; ++j){if(!visited[j] && g[index][j]!=MAXN && dis[j]>dis[index]+g[index][j]){dis[j]=dis[index]+g[index][j];}}}}//有向边转置，其实就是矩阵转置void Traverse(){int i, j;for(i=1; i<=n; ++i){for(j=1; j<=i; ++j){int temp = g[i][j];g[i][j] = g[j][i];g[j][i] = temp;}}}int main(){int i, j;while(cin>>n>>m>>x){//读数据并建图for(i=1; i<=n; ++i)for(j=1; j<=n; ++j)if(i == j)g[i][j] = 0;elseg[i][j] = MAXN;while(m--){int a, b, t;cin>>a>>b>>t;g[a][b] = t;}Dijkistra(dis1);Traverse();Dijkistra(dis2);int nMax = 0;for(i=1; i<=n; ++i){if(i != x){if(nMax<dis1[i]+dis2[i] && dis1[i]+dis2[i]<MAXN)nMax = dis1[i] + dis2[i];}}cout<<nMax<<endl;}return 0;}