hdu 3986 Harry Potter and the Final Battle

一个水题WA了60发，数组没开大，这OJ也不提示RE，光提示WA。。。。。。

思路：先求出最短路，如果删除的边不是最短路上的，那么对结果没有影响，要有影响，只能删除最短路上的边。所以枚举一下最短路上的边，每次求最短路即可。

#include<stdio.h>
#include<vector>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn = 1000 + 50;
const int INF = 0x7FFFFFFF;
int n, m, anss;
vector<int>ljb[maxn];
vector<int>bbb[maxn][maxn];
int jz[maxn][maxn];
int cost[50000 + 50], flag[50000 + 50], ff[maxn], dist[maxn];
int s[maxn], path[maxn];

void SPFA()
{
    int i, ii;
    queue<int>Q;
    memset(ff, 0, sizeof(ff));
    for (i = 0; i <= n; i++) dist[i] = INF, s[i] = -1;
    ff[1] = 1; Q.push(1); dist[1] = 0; s[1] = 1;
    while (!Q.empty())
    {
        int h = Q.front(); Q.pop(); ff[h] = 0;
        for (i = 0; i < ljb[h].size(); i++)
        {
            for (ii = 0; ii < bbb[h][ljb[h][i]].size(); ii++)
            {
                if (flag[bbb[h][ljb[h][i]][ii]] ==0)
                {
                    if (dist[h] + cost[bbb[h][ljb[h][i]][ii]] < dist[ljb[h][i]])
                    {
                        dist[ljb[h][i]] = dist[h] + cost[bbb[h][ljb[h][i]][ii]];
                        s[ljb[h][i]] = h;
                        if (ff[ljb[h][i]] == 0)
                        {
                            ff[ljb[h][i]] = 1;
                            Q.push(ljb[h][i]);
                        }
                    }
                }
            }
        }
    }
}

int main()
{
    int sb;
    scanf("%d", &sb);
    while (sb--)
    {
        int i, j, u, v;
        scanf("%d%d", &n, &m);
        memset(flag, 0, sizeof(flag));
        for (i = 0; i <= n; i++) for (j = 0; j <= n; j++) jz[i][j] = INF;
        for (i = 0; i <= n; i++) for (j = 0; j <= n; j++) bbb[i][j].clear();
        for (i = 0; i <= n; i++) ljb[i].clear();
        for (i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &u, &v, &cost[i]);
            if (jz[u][v] == INF) jz[u][v] = i, jz[v][u] = i;
            if (cost[i] < cost[jz[u][v]]) jz[u][v] = i, jz[v][u] = i;
            bbb[u][v].push_back(i);
            bbb[v][u].push_back(i);
            ljb[u].push_back(v);
            ljb[v].push_back(u);
        }
        SPFA();
        anss = -1;
        if (dist[n] != INF)
        {
            path[1] = n; int q = 2;
            while (1)
            {
                path[q] = s[path[q - 1]];
                if (path[q] == 1) break;
                q++;
            }
            for (i = 1; i <= q - 1; i++)
            {
                flag[jz[path[i]][path[i + 1]]] = 1;
                SPFA();
                if (dist[n] == INF){ anss = -1; break; }
                if (dist[n] > anss) anss = dist[n];
                flag[jz[path[i]][path[i + 1]]] = 0;
            }
        }
        printf("%d
", anss);
    }
    return 0;
}