The Battle of Guandu
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
无
Description
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 wi. Some of the battlefields with wi=2 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 wi=1 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 wi=0 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
Output
one line containing Case #x:, where x is the test case number (starting
from 1). Then output 4 lines with 4 characters each. indicate the
recovered board.
Sample Input
2
2 3
2 3
1 1
1 1
0 1 2
1 1
1
1
1
2
Sample Output
Case #1: 1
Case #2: -1
HINT
题意
在一场战争中，有m(<=100000)个战场和n(<=100000)个村 庄，每个战场有一个重要度，重要度为0表示在这个战场己方输赢无所谓，重要度为1表示己方不能输，重要度为2表示己方必须胜出，己方获得战争的最终胜利当 且仅当己方在每个战场的战果均不违背其重要度，每个战场输赢的判据只有人数，人多的一方胜出，若两方人数相同则打平，对于第i个村庄，每花费c[i]的代 价能够征用一个人派到己方x[i]战场，同时有一个人会跑到敌方y[i]战场，问己方能否获得战争的最终胜利，若能，求出最小代价。
题解：
建图跑spfa就好，如果你考虑每个战场我方人数-敌方人数的数量的话，整个战场的人数实际上是守恒的。你可以花费c[i]代价，使得x[i]加一个人，y[i]减一个人。那么我们就建边跑最短路就好啦，让不重要的战场当成源点，重要的战场当成你要求的最短路就好了
d[i]表示从不重要战场拉一个人到i战场需要的最小代价，只要拉一个人就会胜利
代码：
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<vector>
using namespace std;
#define maxn 100005
long long inf = 999999999999999LL;
int x[maxn];
int y[maxn];
long long c[maxn];
int p[maxn];
int vis[maxn];
long long d[maxn];
vector<pair<int,long long> >G[maxn];
int main()
{
int t;scanf("%d",&t);
for(int cas = ;cas <= t;cas++)
{
int n,m;scanf("%d%d",&n,&m);
for(int i=;i<=m;i++)
G[i].clear();
for(int i=;i<=n;i++)
scanf("%d",&x[i]);
for(int i=;i<=n;i++)
scanf("%d",&y[i]);
for(int i=;i<=n;i++)
scanf("%lld",&c[i]);
for(int i=;i<=m;i++)
scanf("%d",&p[i]); for(int i=;i<=n;i++)
{
if(p[x[i]]==)continue;
G[y[i]].push_back(make_pair(x[i],c[i]));
} queue<int> Q;
for(int i=;i<=m;i++)
{
if(p[i]==)
{
d[i]=;
vis[i]=;
Q.push(i);
}
else
{
d[i]=inf;
vis[i]=;
}
} while(!Q.empty())
{
int now = Q.front();
Q.pop();
vis[now]=;
for(int i=;i<G[now].size();i++)
{
int v = G[now][i].first;
if(d[v]>G[now][i].second + d[now])
{
d[v] = G[now][i].second + d[now];
if(vis[v])continue;
Q.push(v);
vis[v]=;
}
}
} long long ans = ;
int flag = ;
for(int i=;i<=m;i++)
{
if(p[i]==)
{
if(d[i]==inf)
{
flag = ;
break;
}
ans += d[i];
}
}
if(flag)
printf("Case #%d: -1\n",cas);
else
printf("Case #%d: %lld\n",cas,ans);
}
}