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hdu 3986 Harry Potter and the Final Battle

2015-05-27 16:51 471 查看
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
!= 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
== INF){ anss = -1; break; }
if (dist
> anss) anss = dist
;
flag[jz[path[i]][path[i + 1]]] = 0;
}
}
printf("%d\n", anss);
}
return 0;
}
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