首页 > 代码库 > [swustoj1739] 魔术球问题 (最大流,最小路径覆盖)
[swustoj1739] 魔术球问题 (最大流,最小路径覆盖)
题目链接:https://www.oj.swust.edu.cn/problem/show/1739
从1开始枚举球的个数,每次从残余网络更新总流量,最小路径覆盖刚好大于n时ret-1便是最多球。
之后根据容量为0的边找回匹配边即可。
用x << 1和x << 1 | 1拆点 比较方便。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef struct Edge { 5 int u, v, w, next; 6 }Edge; 7 8 const int inf = 0x7f7f7f7f; 9 const int maxn = 100200; 10 11 int cnt, dhead[maxn]; 12 int cur[maxn], dd[maxn]; 13 Edge dedge[maxn<<1]; 14 // bool vis[maxn]; // 记录经过的点 15 int S, T, N; 16 17 void init() { 18 memset(dhead, -1, sizeof(dhead)); 19 for(int i = 0; i < maxn; i++) dedge[i].next = -1; 20 S = 0; cnt = 0; 21 } 22 23 void adde(int u, int v, int w, int c1=0) { 24 dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; 25 dedge[cnt].next = dhead[u]; dhead[u] = cnt++; 26 dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; 27 dedge[cnt].next = dhead[v]; dhead[v] = cnt++; 28 } 29 30 bool bfs(int s, int t, int n) { 31 // memset(vis, 0, sizeof(vis)); 32 queue<int> q; 33 for(int i = 0; i < n; i++) dd[i] = inf; 34 dd[s] = 0; 35 q.push(s); 36 while(!q.empty()) { 37 int u = q.front(); q.pop(); 38 for(int i = dhead[u]; ~i; i = dedge[i].next) { 39 if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) { 40 dd[dedge[i].v] = dd[u] + 1; 41 // vis[dedge[i].v] = 1; 42 if(dedge[i].v == t) return 1; 43 q.push(dedge[i].v); 44 } 45 } 46 } 47 return 0; 48 } 49 50 int dinic(int s, int t, int n) { 51 int st[maxn], top; 52 int u; 53 int flow = 0; 54 while(bfs(s, t, n)) { 55 for(int i = 0; i < n; i++) cur[i] = dhead[i]; 56 u = s; top = 0; 57 while(cur[s] != -1) { 58 if(u == t) { 59 int tp = inf; 60 for(int i = top - 1; i >= 0; i--) { 61 tp = min(tp, dedge[st[i]].w); 62 } 63 flow += tp; 64 for(int i = top - 1; i >= 0; i--) { 65 dedge[st[i]].w -= tp; 66 dedge[st[i] ^ 1].w += tp; 67 if(dedge[st[i]].w == 0) top = i; 68 } 69 u = dedge[st[top]].u; 70 } 71 else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) { 72 st[top++] = cur[u]; 73 u = dedge[cur[u]].v; 74 } 75 else { 76 while(u != s && cur[u] == -1) { 77 u = dedge[st[--top]].u; 78 } 79 cur[u] = dedge[cur[u]].next; 80 } 81 } 82 } 83 return flow; 84 } 85 86 int k, n; 87 int path[maxn]; 88 bool vis[maxn]; 89 90 bool ok(int x) { 91 int y = (int)sqrt(x); 92 return x == y * y; 93 } 94 95 int main() { 96 // freopen("in", "r", stdin); 97 while(~scanf("%d", &n)) { 98 init(); 99 S = 0, T = 10001, N = T + 1; 100 int ret = 0; 101 int flow = 0; 102 while(1) { 103 ret++; 104 for(int i = 1; i < ret; i++) { 105 if(ok(i+ret)) adde(i<<1, (ret<<1)|1, 1); 106 } 107 adde(S, ret<<1, 1); 108 adde((ret<<1)|1, T, 1); 109 flow += dinic(S, T, N); 110 if(ret - flow > n) break; 111 } 112 memset(vis, 0, sizeof(vis)); 113 memset(path, -1, sizeof(path)); 114 printf("%d\n", ret-1); 115 int q = 0; 116 for(int i = 1; i < ret; i++) { 117 for(int j = dhead[i<<1]; ~j; j=dedge[j].next) { 118 if(!dedge[j].w) { 119 path[i] = dedge[j].v >> 1; 120 break; 121 } 122 } 123 } 124 for(int i = 1; i < ret; i++) { 125 if(!vis[i]) { 126 vis[i] = 1; 127 printf("%d", i); 128 int j = path[i]; 129 while(j != 0 && j != T) { 130 printf(" %d", j); 131 vis[j] = 1; 132 j = path[j]; 133 } 134 printf("\n"); 135 } 136 } 137 } 138 return 0; 139 }
[swustoj1739] 魔术球问题 (最大流,最小路径覆盖)
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