題意：
給一張 $N+1$點 $M$

邊無向圖，每個點有一個權值 $Val$。
現在要求從 $0$

點開多隻坦克，停在某些點使得覆蓋點 $\sum Va{l}_{sel}$

e c t e d ≥ ∑ V a l t o t a l 2 \sum Val_{selected}\ge\dfrac{\sum Val_{total}}{2}
求所有坦克開過的距離和的最小值。
$N\le10^2,M\le10^4,dis_{i\in[1,M]}\le10^2,val_{i\in[1,N]}\le10^2$

還算勉強的拼題吧
跑一遍floyd然後01揹包 dij也行啦
當然跑dij,spfa之類的也可以
一開始揹包搞反了還行8

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define add_edge(a,b,c) nxt[++tot]=head[a],head[a]=tot,to[tot]=b,val[tot]=c
#define PII pair<int,int>
priority_queue<PII,vector<PII>,greater<PII> >Q;
int T,N,M,tot=0,sum=0,ans=0x3f3f3f3f;
int head[105]={},nxt[20005]={},to[20005]={},val[20005]={};
int id[105][105]={};
int f[2000005]={};
int dis[105]={},power[105]={};
bool vis[105]={};
void dijkstra()
{
    Q.push(make_pair(0,0));
    while(!Q.empty())
	{
        int x=Q.top().second; Q.pop(); vis[x]=1;
        for(int i=head[x];i;i=nxt[i])
        {
            if(dis[to[i]]>dis[x]+val[i])
            {
                dis[to[i]]=dis[x]+val[i];
                if(!vis[to[i]])Q.push(make_pair(dis[to[i]],to[i]));
            }
        }
    }
}
int main() {
    scanf("%d",&T);
    while(T--)
	{
        scanf("%d%d",&N,&M); 
        memset(head,0,sizeof(head));
   		memset(vis,0,sizeof(vis));
   		memset(dis,0x3f,sizeof(dis));
        for(int i=1;i<=2000000;++i)f[i]=0x3f3f3f3f;
		memset(id,0,sizeof(id));
		sum=tot=f[0]=dis[0]=0; ans=0x3f3f3f3f;
        for(int a,b,c,i=1;i<=M;++i)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(a==b)continue;
            if(id[a][b])val[id[a][b]]=min(val[id[a][b]],c); else add_edge(a,b,c),id[a][b]=tot;
            if(id[b][a])val[id[b][a]]=min(val[id[b][a]],c); else add_edge(b,a,c),id[b][a]=tot;
        }
        for(int i=1;i<=N;++i)scanf("%d",&power[i]),sum+=power[i];
        dijkstra();
        for(int i=1;i<=N;++i)
        for(int j=sum;j>=power[i];--j)
        {
        	if(!vis[i])continue; //!!!!!!!!!!!!!
        	f[j]=min(f[j],f[j-power[i]]+dis[i]);
        }
        for(int i=sum;i>sum/2;--i)ans=min(ans,f[i]);
        if(ans==0x3f3f3f3f)printf("impossible\n");
        else printf("%d\n",ans);
    }
    return 0;
}   
        for(int i=1;i<=N;++i)
        for(int j=sum;j>=power[i];--j)
        f[j]=min(f[j],f[j-power[i]]+dis[i]);
        
        ans=0x3f3f3f3f;
        for(int i=sum;i>(sum>>1);--i)ans=min(ans,f[i]);
        if(ans==0x3f3f3f3f)puts("impossible");
        else printf("%d\n",ans);
    }
    return 0;
}

data generator
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<cctype>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<queue>
#include<sstream>
using namespace std;
#define ll long long
ll GenRand(const ll Lim1,ll Lim2)
{
	++Lim2;
	ll ret=Lim1;
	int t=0;
	while(t<100)
	{
		if(rand()/(RAND_MAX+1.0)<0.1)break;
		ret+=rand();
		ret%=Lim2;
		++t;
	}
	while(ret<Lim1)ret+=Lim1;
	ret%=Lim2;
	return ret;
}
int N,M;
stringstream ss;

int main( int argc, char *argv[] )
{ 
	freopen("data.in","w",stdout);
    int seed=time(NULL);
    if(argc > 1)//如果有引數
    {
        ss.clear();
        ss<<argv[1];
        ss>>seed;//把引數轉換成整數賦值給seed
    }
    srand(seed);
	
	printf("25\n");
	for(int i=1;i<=25;++i)
	{
		N=GenRand(1,100),M=GenRand(1,10000);
		printf("%d %d\n",N,M);
		for(int a,b,c,i=1;i<=M;++i)
		{
			a=GenRand(0,100);
			b=GenRand(0,100);
			while(a==b)b=GenRand(0,100);
			c=GenRand(0,100);
			printf("%d %d %d\n",a,b,c);
		}
		for(int i=1;i<=N;++i)printf("%d\n",GenRand(1,100));
	}
	
	return 0;
}

對拍bat↓

@echo off

:loop
	data_generator.exe %random% > data.in
	std.exe < data.in > std.out
	my.exe < data.in > my.out

	fc my.out std.out

if not errorlevel 1 goto loop
pause

goto loop