标签：hdu 3339 in action   最短路   dp   

In Action

Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team in the US, the number of nuclear weapons have soared across the globe.
Nowadays,the crazy boy in FZU named AekdyCoin possesses some nuclear weapons and wanna destroy our world. Fortunately, our mysterious spy-net has gotten his plan. Now, we need to stop it.
But the arduous task is obviously not easy. First of all, we know that the operating system of the nuclear weapon consists of some connected electric stations, which forms a huge and complex electric network. Every electric station has its power value. To start the nuclear weapon, it must cost half of the electric network‘s power. So first of all, we need to make more than half of the power diasbled. Our tanks are ready for our action in the base(ID is 0), and we must drive them on the road. As for a electric station, we control them if and only if our tanks stop there. 1 unit distance costs 1 unit oil. And we have enough tanks to use.
Now our commander wants to know the minimal oil cost in this action.

2
2 3
0 2 9
2 1 3
1 0 2
1
3
2 1
2 1 3
1
3

5
impossible

            问控制一半以上的核值，需要最   少的油费？

题解：一开始理解成只要有坦克经过就相当于销毁了一个核电站，硬是搞不出来。看懂题意后不就是最短路+01背包么。。

         先跑最短路求出0到任意点的最短距离，然后背包求出大于一般核值得最小油费。

#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdio>

using namespace std;

const int MAXN=110;
const int INF=0x3f3f3f3f;
bool vis[MAXN];
int lowcost[MAXN],cost[MAXN][MAXN];
int n,m;
void Dijkstra(int beg) {
    for(int i=0; i<n; i++) {
        lowcost[i]=INF;
        vis[i]=false;
    }
    lowcost[beg]=0;
    for(int j=0; j<n; j++) {
        int k=-1;
        int Min=INF;
        for(int i=0; i<n; i++)
            if(!vis[i]&&lowcost[i]<Min) {
                Min=lowcost[i];
                k=i;
            }
        if(k==-1)break;
        vis[k]=true;
        for(int i=0; i<n; i++)
            if(!vis[i]&&lowcost[k]+cost[k][i]<lowcost[i]) {
                lowcost[i]=lowcost[k]+cost[k][i];
            }
    }
}

int val[MAXN];
int dp[MAXN*MAXN];

int main() {
    //freopen("test.in","r",stdin);
    int t;
    cin>>t;
    while(t--) {
        scanf("%d%d",&n,&m);
        n++;
        int u,v,w;
        for(int i=0; i<n; i++)
            for(int j=i; j<n; j++) {
                if(i==j)cost[i][j]=0;
                else  cost[i][j]=cost[j][i]=INF;
            }
        for(int i=1; i<=m; i++) {
            scanf("%d%d%d",&u,&v,&w);
            cost[u][v]=cost[v][u]=min(cost[u][v],w);
        }
        int sum=0;
        for(int i=1; i<n; i++) {
            scanf("%d",&val[i]);
            sum+=val[i];
        }
        Dijkstra(0);
        for(int i=0; i<=sum; i++)dp[i]=INF;
        dp[0]=0;
        for(int i=1; i<n; i++) {
            for(int j=sum; j>=val[i]; j--)
                dp[j]=min(dp[j],dp[j-val[i]]+lowcost[i]);
        }
        int ans=INF;
        for(int i=sum/2+1; i<=sum; i++) {
            ans=min(ans,dp[i]);
        }
        if(ans==INF)printf("impossible\n");
        else        printf("%d\n",ans);
    }
    return 0;
}

HDU 3339 In Action（最短路+DP）

标签：hdu 3339 in action   最短路   dp   

原文地址：http://blog.csdn.net/acm_baihuzi/article/details/47738301