标签：hdu3339

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

题意：开始没整明白题意，题意是在0点有很多坦克，指派一些坦克到一些点去，一个坦克只能控制一个点，问怎样能控制一半以上的能量并使总路径最短。
题解：用SPFA求出所有点到0点的最短距离，然后以总路径为背包，点的能量值为价值，对每个点只有去和不去两种状态，用01背包解决。

#include <stdio.h>
#include <string.h>
#include <queue>
#define inf 0x3f3f3f3f
#define maxn 102
#define maxm 10002
using std::queue;

int head[maxn], id, dist[maxn];
int pow[maxn], dp[100 * maxn];
struct Node{
	int to, cost, next;
} E[maxm << 1];
bool vis[maxn];

int max(int a, int b){
	return a > b ? a : b;
}

void addEdge(int u, int v, int cost)
{
	E[id].to = v; E[id].cost = cost;
	E[id].next = head[u]; head[u] = id++;
	E[id].to = u; E[id].cost = cost;
	E[id].next = head[v]; head[v] = id++;
}

void SPFA(int s, int n, int m)
{
	int i, u, v, tmp;
	for(i = 0; i <= n; ++i){
		vis[i] = false; dist[i] = inf;
	}
	u = s; vis[u] = true; dist[u] = 0;
	queue<int> Q;
	Q.push(u);
	while(!Q.empty()){
		u = Q.front(); Q.pop();
		vis[u] = false;
		for(i = head[u]; i != -1; i = E[i].next){
			tmp = dist[u] + E[i].cost;
			v = E[i].to;
			if(tmp < dist[v]){
				dist[v] = tmp;
				if(!vis[v]){
					vis[v] = true;
					Q.push(v);
				}
			}
		}
	}
}

void zeroOnePack(int n, int sumDist)
{
	int i, j;
	memset(dp, 0, sizeof(int) * (sumDist + 1));
	for(i = 1; i <= n; ++i)
		for(j = sumDist; j >= dist[i]; --j){
			dp[j] = max(dp[j], dp[j-dist[i]] + pow[i]);
		}
}

int main()
{
	int n, m, u, v, cost, i, t, sumDist, sumCost;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &n, &m);
		memset(head, -1, sizeof(int) * (n + 1));
		for(i = id = 0; i < m; ++i){
			scanf("%d%d%d", &u, &v, &cost);
			addEdge(u, v, cost);
		}
		for(i = 1, sumCost = 0; i <= n; ++i){
			scanf("%d", &pow[i]);
			sumCost += pow[i];
		}
		SPFA(0, n, id);
		for(i = sumDist = 0; i <= n; ++i)
			if(dist[i] != inf) sumDist += dist[i];
		zeroOnePack(n, sumDist);
		sumCost >>= 1;
		cost = -1;
		for(i = 1; i <= sumDist; ++i)
			if(dp[i] > sumCost){
				cost = i; 
				printf("%d\n", cost);
				break;
			}
		if(cost == -1) printf("impossible\n");
	}
	return 0;
}

HDU3339 In Action 【SPFA】+【01背包】,布布扣,bubuko.com

HDU3339 In Action 【SPFA】+【01背包】

标签：hdu3339

原文地址：http://blog.csdn.net/chang_mu/article/details/38559491