POJ 3159 Candies

　　题意是给A和B发糖果，B的糖果数 – A的糖果数 <= c, 也就是B <= A + c，最后求n比1最多多

几个糖果。题目只有这一个约束条件，建图不难。将AB看成有向图的边，然后c看成边的权值，转化成

最短路来求解，大牛们都说了SPFA + queue会超时，所以用了SPFA + stack。因为这道题没有负

权的边，也可以用堆优化的dij来求这个最短路。

 

SPFA + Stack

/*Accepted    2396K    532MS    C++    1363B    2012-08-06 15:32:00*/

#include<cstdio>

#include<cstring>

#include<cstdlib>



const int inf = 0x3f3f3f3f;

const int V = 30005;

const int E = 150005;

int pnt[E], cost[E], nxt[E];

int head[V], e, dist[V];

bool vis[V];



int relax( int u, int v, int c)

{

    if( dist[v] > dist[u] + c) {

        dist[v] = dist[u] + c;

        return 1;

    }

    return 0;

}



void addedge( int u, int v, int c)

{

    pnt[e] = v; cost[e] = c; nxt[e] = head[u]; head[u] = e ++;

}



int SPFA( int src, int n)

{

    int i;

    for(i = 1; i <= n; ++ i)

    {

         vis[i] = false; dist[i] = inf;

    }

    dist[src] = 0;

    int S[E], top = 1;

    S[0] = src; vis[src] = true;

    while(top) {

        int u, v;

        u = S[ --top]; vis[u] = false;

        for(i = head[u]; i != -1; i = nxt[i]) {

            v = pnt[i];

            if( 1 == relax( u, v, cost[i]) && !vis[v]) {

                S[ top ++] = v; vis[v] = true;

            }

        }

    }

    return dist[n];

}



int main()

{

    int n, m;

    while( scanf( "%d%d", &n, &m) == 2)

    {

        int a, b, c;

        e = 0;

        memset( vis, false, sizeof vis);

        memset( head, -1, sizeof head);

        while( m --)

        {

            scanf( "%d%d%d", &a, &b, &c);

            addedge( a, b, c);

        }

        printf( "%d\n", SPFA( 1, n));

    }

    return 0;

}

 

 Dijkstra

/*Accepted    2392K    610MS    C++    1376B    2012-08-06 15:27:31*/

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<queue>

using namespace std;



const int MAXN = 30030, MAXM = 150150;

const int inf = 0x3f3f3f3f;

int first[MAXN], next[MAXM], v[MAXM], w[MAXM], dist[MAXN];

int n, m, e;

typedef pair<int, int> pii;



void addedge(int a, int b, int c)

{

    v[e] = b, w[e] = c;

    next[e] = first[a], first[a] = e ++;

}



void ReadGraph()

{

    int a, b, c;

    e = 0;

    memset(first, -1, sizeof first);

    while(m --)

    {

        scanf("%d%d%d", &a, &b, &c);

        addedge(a, b, c);

    }

}



int Dijkstra(int src, int n)

{

    int i, x;

    pii u;

    priority_queue<pii, vector<pii>, greater<pii> > q;

    for(i = 1; i <= n; i ++) dist[i] = inf;

    dist[src] = 0;

    q.push(make_pair(dist[src], src));

    while(!q.empty())

    {

        u = q.top(), q.pop();

        x = u.second;

        if(dist[x] != u.first) continue;

        if(n == x) break;

        for(i = first[x]; i != -1; i = next[i])

        {

            if(dist[v[i]] > dist[x] + w[i])

            {

                dist[v[i]] = dist[x] + w[i];

                q.push(make_pair(dist[v[i]], v[i]));

            }

        }

    }

    return dist[n];

}



int main()

{

    while(scanf("%d%d", &n, &m) == 2)

    {

        ReadGraph();

        printf("%d\n", Dijkstra(1, n));

    }

    return 0;

}