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#include #include #include #include using namespace std;const int maxn = 1000 + 50;const int INF = 0x7FFFFFFF;int n, m, anss;vector ljb[maxn];vector 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 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\n", anss); } return 0;}