HDU_3339 In Action(Dijkstra + DP)

　　个人感觉这个题挺好，用到了 最短路+DP，题意很容易就看出来了，其实就是0-1背包问题，状态转移方程：dp[j] = min(dp[j], dp[j-v[i]] + w[i]), 背包容量V = sum(v[0]+v[1] +...+v[n]}，]其中每点的power为v[i], 0点到i点的最短路为w[i]；然后再在i = [sum/2+1, sum]范围内找min(dp[i])。

　　ps：偶悲剧的把i的范围写成i = [sum/2, sum]了，贡献无数WA。。。郁闷！！！

My Code：

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

using namespace std;

const int N = 105;
const int inf = 0x6ffffff;

int dis[N][N];
int dp[N*N];
int vis[N];
int low[N];
int v[N];

int Dijkstra(int n)
{
    int flag, i, j, min;
    for(i = 0; i <= n; i++)
    {
        low[i] = dis[0][i];
        vis[i] = 0;
    }
    vis[0] = 1;
    for(i = 0; i < n; i++)
    {
        min = inf; flag = -1;
        for(j = 0; j <= n; j++)
        {
            if(min > low[j] && !vis[j])
            {
                min = low[j];
                flag = j;
            }
        }
        if(flag == -1)
            break;
        vis[flag] = 1;
        for(j = 0; j <= n; j++)
        {
            if(!vis[j] && dis[flag][j] < inf)
            {
                if(low[j] > dis[flag][j] + low[flag])
                    low[j] = dis[flag][j] + low[flag];
            }
        }
    }
    if(flag != -1)
        return 1;
    return 0;
}

int main()
{
    //freopen("data.in", "r", stdin);

    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, m, i, j, a, b, c, sum = 0;
        scanf("%d%d", &n, &m);
        for(i = 0; i <= n; i++)
            for(j = 0; j <= n; j++)
                if(i == j)
                    dis[i][j] = 0;
                else
                    dis[i][j] = inf;
        while(m--)
        {
            scanf("%d%d%d", &a, &b, &c);
            if(c < dis[a][b])
                dis[a][b] = dis[b][a] = c;
        }

        v[0] = 0;
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &v[i]);
            sum += v[i];
        }

        if(!Dijkstra(n))
        {
            printf("impossible\n");
            continue;
        }

        for(i = 1; i <= sum; i++)
            dp[i] = inf;

        dp[0] = 0;
        for(i = 0; i <= n; i++)
            for(j = sum; j >= v[i]; j--)
                if(dp[j] > dp[j-v[i]]+low[i])
                    dp[j] = dp[j-v[i]]+low[i];
        int min = inf;
        for(i = sum/2+1; i <= sum; i++)
            if(min > dp[i])
                min = dp[i];
        printf("%d\n", min);
    }
    return 0;
}