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最短路+二分。
对容量进行二分,因为容量和时间是单调关系的,容量越多,能用的边越少,时间会不变或者增加。
因为直接暴力一个一个容量去算会TLE,所以采用二分。
#include<cstdio> #include<vector> #include<cstring> #include<queue> #include<map> #include<algorithm> using namespace std; const int maxn = 10001; const int INF = 0x7FFFFFFF; struct aaa { int u, v, cc, tt; }node[500005]; vector<aaa>ljb[maxn]; int c[500005], dist[maxn], flag[maxn]; int n, m, t; void spfa(int xianzhi) { int iii; queue<int>Q; memset(flag, 0, sizeof(flag)); for (iii = 0; iii<=n; iii++) dist[iii] = INF; dist[1] = 0; Q.push(1); flag[1] = 1; while (!Q.empty()) { int h = Q.front(); Q.pop(); flag[h] = 0; for (iii = 0; iii<ljb[h].size(); iii++) { aaa u = ljb[h][iii]; if (u.cc >= xianzhi) { if (u.u == h) { if (dist[u.u] + u.tt <= dist[u.v]) { dist[u.v] = dist[u.u] + u.tt; if (flag[u.v] == 0) { Q.push(u.v); flag[u.v] = 1; } } } else if (u.v == h) { if (dist[u.v] + u.tt <= dist[u.u]) { dist[u.u] = dist[u.v] + u.tt; if (flag[u.u] == 0) { Q.push(u.u); flag[u.u] = 1; } } } } } } } int main() { int X; scanf("%d", &X); while (X--) { int i, j, u, v, tt; scanf("%d%d%d", &n, &m, &t); for (i = 0; i <= n; i++) ljb[i].clear(); for (i = 0; i <= m; i++) node[i].tt = INF; for (i = 0; i < m; i++) { scanf("%d%d%d%d", &u, &v, &c[i], &tt); node[i].u = u; node[i].v = v; node[i].cc = c[i]; node[i].tt = tt; ljb[v].push_back(node[i]); ljb[u].push_back(node[i]); } sort(c, c + m); int anss = -1, xx, dd, zz; xx = 0; dd = m - 1; zz = (xx + dd) / 2; while (1) { spfa(c[zz]); if (dist[n] <= t) { anss = zz; xx = zz + 1; zz = (xx + dd) / 2; } else { dd = zz; zz = (xx + dd) / 2; } if (xx >= dd) break; } printf("%d\n", c[anss]); } return 0; }
hdu 1839 Delay Constrained Maximum Capacity Path
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原文地址:http://www.cnblogs.com/zufezzt/p/4524398.html