HDU 5545 The Battle of Guandu（最短路）

最新推荐文章于 2019-01-20 18:20:27 发布

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本文探讨了一个基于图论的问题，曹操如何在不同战场间调动士兵以最小化成本，并确保在关键战场上占据优势。通过构建图模型并运用最短路径算法找到最优解。

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思路：首先每次曹操从一个村庄调用一个兵到战场xi时，其实相当于同时往yi战场减少一个兵，如果yi战场的价值度是0的话那么是无所谓的，如果价值是2的话，那么这减少的一个兵你就要通过其他村庄转移来填补这一个减少的兵，否则就会输掉，那么这样一直转移到一个无用的战场，那么就可以满足条件了，所以按照这样建图然后跑一次最短路那么就是最少费用了（一开始以为是网络流什么的...大概数据范围也透露了一点...）

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 100000+500;
int n,m;
#define INF 1LL<<38
int x[maxn],y[maxn],p[maxn],inq[maxn];
LL c[maxn],d[maxn];
vector<pair<int,LL> >e[maxn];
int main()
{
    int T,cas=1;
	scanf("%d",&T);
	while(T--)
	{
		memset(d,0,sizeof(d));
        scanf("%d%d",&n,&m);
		for(int i = 0;i<=n;i++)
			e[i].clear();
		for(int i = 1;i<=n;i++)
			scanf("%d",&x[i]);
		for(int i = 1;i<=n;i++)
			scanf("%d",&y[i]);
        for(int i = 1;i<=n;i++)
			scanf("%lld",&c[i]);
		for(int i = 1;i<=m;i++)
			scanf("%d",&p[i]);
		for(int i = 1;i<=n;i++)
		{
            if(p[x[i]]==0)continue;
			e[y[i]].push_back(make_pair(x[i],c[i]));
		}
		queue<int>q;
		for(int i = 1;i<=m;i++)
		{
			if(p[i]==0)
			{
				q.push(i);
				inq[i]=1;
                d[i]=0;
			}
			else
			{
				d[i]=INF;
				inq[i]=0;
			}
		}
		while(!q.empty())
		{
			int u = q.front();
			q.pop();
			inq[u]=0;
			for(int i = 0;i<e[u].size();i++)
			{
				int v = e[u][i].first;
				if(d[v]>d[u]+e[u][i].second)
				{
					d[v]=d[u]+e[u][i].second;
					if(inq[v])
						continue;
					inq[v]=1;
					q.push(v);
				}
			}
		}
		LL ans = 0;
		int flag = 0;
		for(int i = 1;i<=m;i++)
		{
			if(p[i]==2)
			{
				if(d[i]==INF)
				{
					flag = 1;
					break;
				}
				ans+=d[i];
			}
		}
		printf("Case #%d: ",cas++);
		if(flag)
			printf("-1\n");
		else
			printf("%lld\n",ans);
	}
}

Description

In the year of 200, two generals whose names are Cao Cao and Shao Yuan are fighting in Guandu. The battle of Guandu was a great battle and the two armies were fighting at

different battlefields whose numbers were 1 to

. There were also

villages nearby numbered from 1 to

. Cao Cao could train some warriors from those villages to strengthen his military. For village

, Cao Cao could only call for some number of warriors join the battlefield

. However, Shao Yuan's power was extremely strong at that time. So in order to protect themselves, village

would also send equal number of warriors to battlefield

and join the Yuan Shao's Army. If Cao Cao had called for one warrior from village

, he would have to pay

units of money for the village. There was no need for Cao Cao to pay for the warriors who would join Shao Yuan's army. At the beginning, there were no warriors of both sides in every battlefield.

As one of greatest strategist at that time, Cao Cao was considering how to beat Shao Yuan. As we can image, the battlefields would have different level of importance

. Some of the battlefields with

were very important, so Cao Cao had to guarantee that in these battlefields, the number of his warriors was greater than Shao Yuan's. And some of the battlefields with

were not as important as before, so Cao Cao had to make sure that the number of his warriors was greater or equal to Shao Yuan's. The other battlefields with

had no importance, so there were no restriction about the number of warriors in those battlefields. Now, given such conditions, could you help Cao Cao find the least number of money he had to pay to win the battlefield?

Input

The first line of the input gives the number of test cases,