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;
}