acdrem1083 人民城管爱人民 DP

来源：互联网发布：win10网络共享文件夹编辑：程序博客网时间：2024/04/20 12:24

思路：d(i, 0)表示从节点i到达大运村的最短路径，d(i, 1)表示从节点i到达大运村的次短路径。

1.最短路：当做DAG处理即可。

2.次短路：假设当前在u点处，下一个节点是v。v到终点的最短路是d(v, 0)，次短路是d(v, 1)，u到v的距离是w(u, v)，只能封锁一条路，即只能删除一条边。分成两种情况：

a.将封锁用在了v点到达终点的路上，则此时GG拥有主动权，可以选择最小的路径，即Imin = min(w(u, v) + d(v, 1))，;

b.将封锁用在u-v这条路径上面，则此时城管拥有主动权，可以封锁最短路径，则GG这能选择次短路径Hmin。

则d(u, 1) = max(Imin, Hmin)

AC代码

#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 10000 + 5;int n, m;int d[maxn][2], in[maxn], Top[maxn], c[maxn*20];struct Edge{int from, to, dist, nex;Edge() {}Edge(int u, int v, int d, int ne):from(u), to(v), dist(d), nex(ne){}}edge[maxn*20];int edgenum, head[maxn];void add_edge(int u, int v, int dis) {edge[edgenum] = Edge(u, v, dis, head[u]);head[u] = edgenum++;}void topsort() { //逆向拓扑序 queue<int>q;for(int i = 0; i < n; ++i) if(!in[i]) q.push(i);int num = 0;while(!q.empty()) {int u = q.front(); q.pop();for(int i = head[u]; i != -1; i = edge[i].nex) if(i&1){int v = edge[i].to;--in[v];if(!in[v]) q.push(v);}Top[num++] = u;}}void solve(int u) {if(u == n-1) {d[u][0] = d[u][1] = 0; return;} //终点int num = 0, Imin = inf;for(int i = head[u]; i != -1; i = edge[i].nex) if(!(i&1)){int v = edge[i].to;//最短路d[u][0] = min(d[u][0], d[v][0] + edge[i].dist);//次短路Imin = min(Imin, d[v][1] + edge[i].dist); // GG掌握主动权 c[num++] = d[v][0] + edge[i].dist; //城管有主动权 }if(num < 2) {d[u][1] = inf; return;} //GG无法到达从u到达终点 int fir = c[0], sec = c[1]; //最小和次小 if(fir > sec) swap(fir, sec);for(int i = 2; i < num; ++i) {if(c[i] <= fir) {sec = fir;fir = c[i];}else sec = min(sec, c[i]);}d[u][1] = max(Imin, sec);if(d[u][1] > inf) d[u][1] = inf;if(d[u][0] > inf) d[u][0] = inf;}int main() {int T;scanf("%d", &T);while(T--) {edgenum = 0;memset(head, -1, sizeof(head));memset(d, inf, sizeof(d));memset(in, 0, sizeof(in));scanf("%d%d", &n, &m);int u, v, dis;for(int i = 0; i < m; ++i) {scanf("%d%d%d", &u, &v, &dis);add_edge(u, v, dis);add_edge(v, u, dis);in[u]++;}topsort();for(int i = 0; i < n; ++i) {solve(Top[i]);}if(d[0][1] >= inf) printf("-1\n");else printf("%d\n", d[0][1]);}return 0;}

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