pku2387 Til the Cows Come Home

http://poj.org/problem?id=2387

图论，最短路

单源最短路，Dijkstra O（N^2）

#include <stdio.h>
#include <string.h>
#define N 1234

const int inf = 1<<25;
int n;
int path[N], vis[N];
int cost[N][N], lowcost[N];

void dijkstra(int beg)
{
    int i, j, min1;
    memset(vis, 0, sizeof(vis));
    vis[beg] = 1;
    for(i=0; i<n; i++)
    {
        lowcost[i] = cost[beg][i];
        path[i] = beg;
    }
    lowcost[beg] = 0;
    path[beg] = -1;
    int pre = beg;
    for(i=1; i<n; i++)
    {
        min1 = inf;
        for(j=0; j<n; j++)
        {
            if(vis[j]==0 && lowcost[pre]+cost[pre][j]<lowcost[j])
            {
                lowcost[j] = lowcost[pre] + cost[pre][j];
                path[j] = pre;
            }
        }
        for(j=0; j<n; j++)
        {
            if(vis[j] == 0 && lowcost[j] < min1)
            {
                min1 = lowcost[j];
                pre = j;
            }
        }
        vis[pre] = 1;
    }
}


int main()
{
    int t, i, j, x, y, len;
    scanf("%d%d", &t, &n);
    for(i=0; i<n; i++)
    {
        for(j=0; j<n; j++)
        {
            cost[i][j] = inf;
        }
    }
    for(i=1; i<=t; i++)
    {
        scanf("%d%d%d", &x, &y, &len);
        if(len < cost[x-1][y-1])
        {
            cost[x-1][y-1] = cost[y-1][x-1] = len;
        }
    }
    dijkstra(0);
    printf("%d\n", lowcost[n-1]);
    return 0;
}

单源最短路，SPFA，O（V*E）

#include <stdio.h>
#include <string.h>
#include <queue>
#define N 1234

using namespace std;

int n, m, src;
vector<pair<int, int> > g[N];
int dist[N];
bool inQue[N];
queue<int> que;

void spfa()
{
    memset(dist, 63, sizeof(dist));
    dist[src] = 0;
    while(!que.empty())
    {
        que.pop();
    }
    que.push(src);
    inQue[src] = true;
    while(!que.empty())
    {
        int u = que.front();
        que.pop();
        for(int i=0; i<g[u].size(); i++)
        {
            if(dist[u] + g[u][i].second < dist[g[u][i].first])
            {
                dist[g[u][i].first] = dist[u] + g[u][i].second;
                if(!inQue[g[u][i].first])
                {
                    inQue[g[u][i].first] = true;
                    que.push(g[u][i].first);
                }
            }
        }
        inQue[u] = false;
    }
}

int main()
{
    int t, i, j, x, y, len;
    pair<int, int> pair1;
    scanf("%d%d", &t, &n);
    for(i=1; i<=t; i++)
    {
        scanf("%d%d%d", &x, &y, &len);
        g[x].push_back(make_pair(y, len));
        g[y].push_back(make_pair(x, len));
    }
    src = 1;
    spfa();
    printf("%d\n", dist[n]);
    return 0;
}

floyd求图中任意两点间距离 O(N^3)，对于本题会TLE

#include <stdio.h>
#include <string.h>
#define N 1234

int n, g[N][N];
const int inf = 1<<29;

int min(int x, int y)
{
    return x<y? x: y; 
}

void floyd()
{
    int i, j, k;
    for(k=1; k<=n; k++)
    {
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                g[i][j] = min(g[i][j], g[i][k]+g[k][j]);
            }
        }
    }
}

int main()
{
    int t, i, j, x, y, len;
    scanf("%d%d", &t, &n);
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=n; j++)
        {
            g[i][j] = inf;
        }
    }
    for(i=1; i<=t; i++)
    {
        scanf("%d%d%d", &x, &y, &len);
        if(len < g[x][y])
        {
            g[x][y] = g[y][x] = len;
        }
    }
    floyd();
    printf("%d\n", g[n][1]);
    return 0;
}