3931: [CQOI2015]网络吞吐量
Time Limit: 10 Sec Memory Limit: 512 MBSubmit: 1555 Solved: 637
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Description
路由是指通过计算机网络把信息从源地址传输到目的地址的活动，也是计算机网络设计中的重点和难点。网络中实现路由转发的硬件设备称为路由器。为了使数据包最快的到达目的地，路由器需要选择最优的路径转发数据包。例如在常用的路由算法OSPF(开放式最短路径优先)中，路由器会使用经典的Dijkstra算法计算最短路径，然后尽量沿最短路径转发数据包。现在，若已知一个计算机网络中各路由器间的连接情况，以及各个路由器的最大吞吐量（即每秒能转发的数据包数量），假设所有数据包一定沿最短路径转发，试计算从路由器1到路由器n的网络的最大吞吐量。计算中忽略转发及传输的时间开销，不考虑链路的带宽限制，即认为数据包可以瞬间通过网络。路由器1到路由器n作为起点和终点，自身的吞吐量不用考虑，网络上也不存在将1和n直接相连的链路。
Input
输入文件第一行包含两个空格分开的正整数n和m，分别表示路由器数量和链路的数量。网络中的路由器使用1到n编号。接下来m行，每行包含三个空格分开的正整数a、b和d，表示从路由器a到路由器b存在一条距离为d的双向链路。 接下来n行，每行包含一个正整数c，分别给出每一个路由器的吞吐量。
Output
输出一个整数，为题目所求吞吐量。
Sample Input
7 10
1 2 2
1 5 2
2 4 1
2 3 3
3 7 1
4 5 4
4 3 1
4 6 1
5 6 2
6 7 1
1
100
20
50
20
60
1
1 2 2
1 5 2
2 4 1
2 3 3
3 7 1
4 5 4
4 3 1
4 6 1
5 6 2
6 7 1
1
100
20
50
20
60
1
Sample Output
70
HINT
对于100%的数据，n≤500，m≤100000，d,c≤10^9
Source
分别从1和n点做单源最短路，即可求出哪些边出现在了从1到n的最短路上（最短路不一定唯一）。将这些边加入网络流中，对于原本的点拆点，入点向出点连限制吞吐量容量的边，跑最大流。注意Int64。
1 #include <cstdio> 2 #include <cstring> 3 4 #define int long long 5 6 inline int nextChar(void) { 7 const int siz = 1024; 8 static char buf[siz]; 9 static char *hd = buf + siz; 10 static char *tl = buf + siz; 11 if (hd == tl) 12 fread(hd = buf, 1, siz, stdin); 13 return *hd++; 14 } 15 16 inline int nextInt(void) { 17 register int ret = 0; 18 register int neg = false; 19 register int bit = nextChar(); 20 for (; bit < 48; bit = nextChar()) 21 if (bit == '-')neg ^= true; 22 for (; bit > 47; bit = nextChar()) 23 ret = ret * 10 + bit - 48; 24 return neg ? -ret : ret; 25 } 26 27 const int inf = 2e18; 28 const int siz = 1000005; 29 30 int n, m; 31 32 struct edge { 33 int x, y, w; 34 }e[siz]; 35 36 int lim[siz]; 37 38 namespace shortestPath 39 { 40 int dis[2][siz]; 41 42 int edges; 43 int hd[siz]; 44 int to[siz]; 45 int nt[siz]; 46 int vl[siz]; 47 48 inline void add(int u, int v, int w) { 49 nt[edges] = hd[u]; to[edges] = v; vl[edges] = w; hd[u] = edges++; 50 nt[edges] = hd[v]; to[edges] = u; vl[edges] = w; hd[v] = edges++; 51 } 52 53 inline void spfa(int *d, int s) { 54 static int que[siz]; 55 static int inq[siz]; 56 static int head, tail; 57 memset(inq, 0, sizeof(inq)); 58 for (int i = 0; i < siz; ++i)d[i] = inf; 59 inq[que[head = d[s] = 0] = s] = tail = 1; 60 while (head != tail) { 61 int u = que[head++], v; inq[u] = 0; 62 for (int i = hd[u]; ~i; i = nt[i]) 63 if (d[v = to[i]] > d[u] + vl[i]) { 64 d[v] = d[u] + vl[i]; 65 if (!inq[v])inq[que[tail++] = v] = 1; 66 } 67 } 68 } 69 70 inline void solve(void) { 71 memset(hd, -1, sizeof(hd)); 72 for (int i = 1; i <= m; ++i) 73 add(e[i].x, e[i].y, e[i].w); 74 spfa(dis[0], 1); 75 spfa(dis[1], n); 76 } 77 } 78 79 namespace networkFlow 80 { 81 int s, t; 82 int edges; 83 int hd[siz]; 84 int to[siz]; 85 int nt[siz]; 86 int fl[siz]; 87 88 inline void add(int u, int v, int f) { 89 nt[edges] = hd[u]; to[edges] = v; fl[edges] = f; hd[u] = edges++; 90 nt[edges] = hd[v]; to[edges] = u; fl[edges] = 0; hd[v] = edges++; 91 } 92 93 int dep[siz]; 94 95 inline bool bfs(void) { 96 static int que[siz], head, tail; 97 memset(dep, 0, sizeof(dep)); 98 dep[que[head = 0] = s] = tail = 1; 99 while (head != tail) { 100 int u = que[head++], v; 101 for (int i = hd[u]; ~i; i = nt[i]) 102 if (fl[i] && !dep[v = to[i]]) 103 dep[que[tail++] = v] = dep[u] + 1; 104 } 105 return dep[t]; 106 } 107 108 int lst[siz]; 109 110 int dfs(int u, int f) { 111 if (u == t || !f)return f; 112 int used = 0, flow, v; 113 for (int i = lst[u]; ~i; i = nt[i]) 114 if (dep[v = to[i]] == dep[u] + 1) { 115 flow = dfs(v, f - used < fl[i] ? f - used : fl[i]); 116 used += flow; 117 fl[i] -= flow; 118 fl[i^1] += flow; 119 if (fl[i])lst[u] = i; 120 if (used == f)return f; 121 } 122 if (!used)dep[u] = 0; 123 return used; 124 } 125 126 inline int maxFlow(void) { 127 int maxFlow = 0, newFlow; 128 while (bfs()) { 129 for (int i = s; i <= t; ++i) 130 lst[i] = hd[i]; 131 while (newFlow = dfs(s, inf)) 132 maxFlow += newFlow; 133 } 134 return maxFlow; 135 } 136 137 inline void solve(void) { 138 s = 0, t = (n + 1) << 1; 139 memset(hd, -1, sizeof(hd)); 140 add(s, 1 << 1, inf); 141 add(n << 1 | 1, t, inf); 142 for (int i = 1; i <= n; ++i) 143 add(i << 1, i << 1 | 1, lim[i]); 144 for (int i = 1; i <= m; ++i) { 145 int x, y, d = shortestPath::dis[0][n]; 146 x = shortestPath::dis[0][e[i].x]; 147 y = shortestPath::dis[1][e[i].y]; 148 if (x + y + e[i].w == d) 149 add(e[i].x << 1 | 1, e[i].y << 1, inf); 150 x = shortestPath::dis[0][e[i].y]; 151 y = shortestPath::dis[1][e[i].x]; 152 if (x + y + e[i].w == d) 153 add(e[i].y << 1 | 1, e[i].x << 1, inf); 154 } 155 printf("%lld ", maxFlow()); 156 } 157 } 158 159 signed main(void) { 160 n = nextInt(); 161 m = nextInt(); 162 for (int i = 1; i <= m; ++i) 163 e[i].x = nextInt(), 164 e[i].y = nextInt(), 165 e[i].w = nextInt(); 166 for (int i = 1; i <= n; ++i) 167 lim[i] = nextInt(); 168 lim[1] = lim[n] = inf; 169 shortestPath::solve(); 170 networkFlow::solve(); 171 }
@Author: YouSiki