hdu 3986 Harry Potter and the Final Battle

时间:2023-03-08 21:22:53

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 = + ;
const int INF = 0x7FFFFFFF;
int n, m, anss;
vector<int>ljb[maxn];
vector<int>bbb[maxn][maxn];
int jz[maxn][maxn];
int cost[ + ], flag[ + ], ff[maxn], dist[maxn];
int s[maxn], path[maxn]; void SPFA()
{
int i, ii;
queue<int>Q;
memset(ff, , sizeof(ff));
for (i = ; i <= n; i++) dist[i] = INF, s[i] = -;
ff[] = ; Q.push(); dist[] = ; s[] = ;
while (!Q.empty())
{
int h = Q.front(); Q.pop(); ff[h] = ;
for (i = ; i < ljb[h].size(); i++)
{
for (ii = ; ii < bbb[h][ljb[h][i]].size(); ii++)
{
if (flag[bbb[h][ljb[h][i]][ii]] ==)
{
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]] == )
{
ff[ljb[h][i]] = ;
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, , sizeof(flag));
for (i = ; i <= n; i++) for (j = ; j <= n; j++) jz[i][j] = INF;
for (i = ; i <= n; i++) for (j = ; j <= n; j++) bbb[i][j].clear();
for (i = ; i <= n; i++) ljb[i].clear();
for (i = ; 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 = -;
if (dist[n] != INF)
{
path[] = n; int q = ;
while ()
{
path[q] = s[path[q - ]];
if (path[q] == ) break;
q++;
}
for (i = ; i <= q - ; i++)
{
flag[jz[path[i]][path[i + ]]] = ;
SPFA();
if (dist[n] == INF){ anss = -; break; }
if (dist[n] > anss) anss = dist[n];
flag[jz[path[i]][path[i + ]]] = ;
}
}
printf("%d\n", anss);
}
return ;
}