HDU 3986 Harry Potter and the Final Battle 删掉任意一条边的最长最短路

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题目来源：HDU 3986 Harry Potter and the Final Battle

题意：哈利波特要从1到n 不过敌人可以破坏一条边求最坏的情况下到达n需要的最短时间不能到达输出-1

思路：和上一题差不多每次枚举最短路上的边然后去掉该条边在做最短路因为是无向的不要忘记把它的反向边也删除

上一题一定有解这题不一定如果去掉一条边后无法到达n那么输出-1 因为敌人足够聪明

#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;const int maxn = 1010;struct edge{int u, v, w;}a[maxn*maxn];int n, m, num;struct HeapNode{int u, d;bool operator < (const HeapNode& rhs) const{return d > rhs.d;}};vector <edge> edges;vector <int> G[maxn];int d[maxn];int p[maxn];bool vis[maxn];void Dijkstra(int s){for(int i = 0; i <= n; i++)d[i] = 999999999;d[1] = 0;memset(vis, false, sizeof(vis));if(s)memset(p, -1, sizeof(p));priority_queue <HeapNode> Q;HeapNode x;x.u = 1;x.d = 0;Q.push(x);while(!Q.empty()){x = Q.top();Q.pop();int u = x.u;if(vis[u])continue;vis[u] = true;for(int i = 0; i < G[u].size(); i++){edge e = edges[G[u][i]];int v = e.v;if(d[v] > x.d + e.w){if(s)p[v] = G[u][i];d[v] = x.d + e.w;HeapNode xx;xx.u = v;xx.d = d[v];Q.push(xx);}}}}void AddEdge(int u, int v, int w){edge x;x.u = u;x.v = v;x.w = w;edges.push_back(x);x.u = v;x.v = u;edges.push_back(x);num = edges.size();G[u].push_back(num-2);G[v].push_back(num-1);}int main(){int T;scanf("%d", &T);while(T--){scanf("%d %d", &n, &m);for(int i = 0; i <= n; i++)G[i].clear();edges.clear();for(int i = 0; i < m; i++){scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].w);AddEdge(a[i].u, a[i].v, a[i].w);}Dijkstra(1);int ans = -1;int pos = p[n];while(pos != -1){//printf("%d %d\n", pos, pos^1);int w1 = edges[pos].w;int w2 = edges[pos^1].w;edges[pos].w = 999999999;edges[pos^1].w = 999999999;Dijkstra(0);edges[pos].w = w1;edges[pos^1].w = w2;pos = p[edges[pos].u];if(d[n] == 999999999){ans = -1;//有一次不能到达 就输出-1 break;}if(ans == -1 || ans < d[n])ans = d[n];}printf("%d\n", ans);}return 0;}

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