标签：columbuss bargain   hdu 3268   最短路   

Columbus’s bargain

1
4
1 4
2 9
3 5
4 13
2
1 2 3
3 4 6

1 3
2 6
3 4
4 10
1

题意：n个物品，每个物品价值为pi，可以用一个玻璃珠和pi-1个金币换的，也可以由其他便宜的物品加上一定量的金币换得，相同价格的物品可以互换。问每个物品通过这样交换若干次后最少要花的金币数是多少。并且有多少物品的actual price等于其他两个物品actual price之和，注意每个物品如果有多个组合形式也只算作一种。

代码：

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 2005;
const int MAXM = 200010;

typedef pair<int,int> P;

struct Edge
{
    int v,len,next;
}edge[MAXM];

int a[MAXN];
int head[MAXN],dist[MAXN];
int num,n,m;
int vis[MAXN];

void init()
{
    num=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w)
{
    edge[num].v=v;
    edge[num].len=w;
    edge[num].next=head[u];
    head[u]=num++;
}

void Dijkstra()
{
    priority_queue<P,vector<P>,greater<P> >Q;
    memset(dist,INF,sizeof(dist));
    Q.push(P(0,0));
    dist[0]=0;
    while (!Q.empty())
    {
        P p=Q.top();
        Q.pop();
        int u=p.second;
        if (dist[u]<p.first) continue;
        for (int i=head[u];~i;i=edge[i].next)
        {
            int v=edge[i].v;
            if (dist[v]>dist[u]+edge[i].len)
            {
                dist[v]=dist[u]+edge[i].len;
                Q.push(P(dist[v],v));
            }
        }
    }
    for (int i=1;i<=n;i++)
        printf("%d %d\n",i,dist[i]);
    int ans=0;
    memset(vis,0,sizeof(vis));      //每种物品有多种组合方式也只算作一种，用vis来标记
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=n;j++)
        {
            if (i==j) continue;
            for (int k=j+1;k<=n;k++)
            {
                if (i==k||j==k) continue;
                if (dist[i]==dist[j]+dist[k]&&!vis[i])
                {
                    ans++;
                    vis[i]=1;
                }
            }
        }
    }
    printf("%d\n",ans);
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,t,u,v,w;
    scanf("%d",&t);
    while (t--)
    {
        init();
        scanf("%d",&n);
        for (i=1;i<=n;i++)
        {
            scanf("%d%d",&u,&a[i]);
            a[i]--;
            addedge(0,u,a[i]);
        }
        for (i=1;i<=n;i++)
        {
            for (j=1;j<=n;j++)
            {
                if (a[i]==a[j])
                    addedge(i,j,0);
            }
        }
        scanf("%d",&m);
        while (m--)
        {
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
        }
        Dijkstra();
    }
    return 0;
}

Columbus’s bargain (hdu 3268 最短路)

标签：columbuss bargain   hdu 3268   最短路   

原文地址：http://blog.csdn.net/u014422052/article/details/47985327