描述
题解
最短路+背包。第一次做将这两种算法组合的题,好题。
要求最少油耗使得系统瘫痪,而瘫痪的要求是控制的能量超过一半,那么前者很容易想到需要先求最短路,但是求过最短路后并不是每一个电厂都要占领,要保证占领的电厂的总能量超过一半并且耗油最少,这就是01背包的问题了~~~
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
/* * 单源最短路径,Dijkstra算法,邻接矩阵形式,复杂度为O(n^2) * 求出源beg到所有点的最短路径,传入图的顶点数和邻接矩阵cost[][] * 返回各点的最短路径lowcost[],路径pre[],pre[i]记录beg到i路径上的父节点,pre[beg] = -1 * 可更改路径权类型,但是权值必须为非负,下标0~n-1 */
const int MAXN = 110;
const int INF = 0x3f3f3f3f; // 表示无穷
bool vis[MAXN];
int pre[MAXN];
void Dijkstra(int cost[][MAXN], int lowcost[], int n, int beg)
{
for (int i = 0; i < n; i++)
{
lowcost[i] = INF;
vis[i] = false;
pre[i] = -1;
}
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];
pre[i] = k;
}
}
}
}
int cost[MAXN][MAXN];
int lowcost[MAXN];
int power[MAXN];
int dp[MAXN * MAXN];
int main()
{
int T;
scanf("%d", &T);
int n, m;
while (T--)
{
memset(cost, 0x3f, sizeof(cost));
scanf("%d%d", &n, &m);
int st, ed, val;
while (m--)
{
scanf("%d%d%d", &st, &ed, &val);
if (val < cost[st][ed])
{
cost[st][ed] = cost[ed][st] = val;
}
}
int sum = 0;
for (int i = 1; i <= n; i++)
{
scanf("%d", &power[i]);
sum += power[i];
}
Dijkstra(cost, lowcost, n + 1, 0);
memset(dp, 0x3f, sizeof(dp));
dp[0] = 0;
for (int i = 1; i <= n; i++)
{
for (int j = sum; j >= power[i]; j--)
{
dp[j] = min(dp[j], dp[j - power[i]] + lowcost[i]);
}
}
int mid = sum / 2 + 1; // 大于一半
int res = INF;
for (int i = mid; i <= sum; i++)
{
if (dp[i] < res)
{
res = dp[i];
}
}
if (res == INF)
{
printf("impossible\n");
}
else
{
printf("%d\n", res);
}
}
return 0;
}