标签：double code node ems read sizeof cti 负环 bug
最近因发现基础不好而狂刷板子，导致bzoj的计划放了一段时间。现在板子搞得差不多了，还剩一个90分的负环和70分的树刨搞不出来，就很慌，希望大家帮我调一下，都是T了。
接下来重心放到bzoj去，按照hzwer的顺序吧。
下面放代码：
90分负环：
// luogu-judger-enable-o2 #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> #include <iomanip> #include <iostream> using namespace std ; const int maxn = 200011,maxm = 200011,inf = 1e9 ; struct node{ int to,val,pre ; }e[2*maxm]; int T,n,m,x,y,val,cnt ; int head[maxn],dist[maxn] ; bool flag ; bool vis[maxn] ; inline void addedge(int x,int y,int v) { e[++cnt] = (node){ y,v,head[x] } ; head[ x ] = cnt ; } inline int read() { char ch = getchar() ; int x = 0 , f = 1 ; while(ch<‘0‘||ch>‘9‘) { if(ch==‘-‘) f = -1 ; ch = getchar() ; } while(ch>=‘0‘&&ch<=‘9‘) { x = x*10+ch-48 ; ch = getchar() ; } return x*f ; } inline void SPFA(int u) { int v ; vis[ u ] = 1 ; for(int i=head[u];i;i = e[ i ].pre ) { v = e[ i ].to ; if( dist[ u ] + e[ i ].val < dist[ v ] ) { dist[ v ] = dist[ u ] + e[ i ].val ; if(vis[ v ]||flag) { flag = 1 ; break ; } SPFA( v ) ; } } vis[ u ] = 0 ; } int main() { T = read() ; while(T--) { flag = 0 ; cnt = 0 ; n = read() ; m = read() ; for(int i=1;i<=n;i++) dist[ i ] = 0,vis[ i ] = 0,head[ i ] = 0 ; //0 for(int i=1;i<=m;i++) { scanf("%d%d%d",&x,&y,&val) ; addedge(x,y,val) ; if(val>=0) addedge(y,x,val) ; } for(int i=1;i<=n;i++) { SPFA( i ) ; if(flag) break ; } if(flag) printf("YE5\n") ; else printf("N0\n") ; } return 0 ; }
70分树刨：
// luogu-judger-enable-o2 #include<iostream> #include<cstdio> #include<cmath> #include<ctime> #include<queue> #include<algorithm> #include<cstring> using namespace std; #define duke(i,a,n) for(int i = a;i <= n;i++) #define lv(i,a,n) for(int i = a;i >= n;i--) #define clean(a) memset(a,0,sizeof(a)) const int INF = 1 << 30; const int N = 200005; typedef long long ll; typedef double db; template <class T> void read(T &x) { char c; bool op = 0; while(c = getchar(), c < ‘0‘ || c > ‘9‘) if(c == ‘-‘) op = 1; x = c - ‘0‘; while(c = getchar(), c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘; if(op) x = -x; } template <class T> void write(T x) { if(x < 0) putchar(‘-‘), x = -x; if(x >= 10) write(x / 10); putchar(‘0‘ + x % 10); } struct edge { int nxt,r; } e[4 * N]; int n,m,r,cnt; int a[N],lst[N],p; int f[N],d[N],siz[N],son[N],rk[N]; int top[N],id[N],tree[4 * N]; int lazy[4 * N],len = 0; //id新编号dfs序 void add(int x,int y) { e[++len].nxt = lst[x]; e[len].r = y; lst[x] = len; } void dfs1(int u,int fa,int depth) { f[u] = fa; d[u] = depth; siz[u] = 1; for(int k = lst[u];k;k = e[k].nxt) { int y = e[k].r; if(y == fa) continue; dfs1(y,u,depth + 1); siz[u] += siz[y]; if(siz[y] > siz[son[u]] || !son[u]) { son[u] = y; } } } void dfs2(int u,int t) { top[u] = t; id[u] = ++cnt; rk[cnt] = u; if(!son[u]) return; dfs2(son[u],t); for(int k = lst[u];k;k = e[k].nxt) { int y = e[k].r; if(y != son[u] && y != f[u]) dfs2(y,y); } } void push_down(int o,int l,int r) { if(lazy[o]) { lazy[o << 1] += lazy[o]; lazy[o << 1] %= p; lazy[o << 1 | 1] += lazy[o]; lazy[o << 1 | 1] %= p; int len = (r - l + 1); tree[o << 1] += lazy[o] * (len - (len >> 1)); tree[o << 1 | 1] += lazy[o] * (len >> 1); tree[o << 1] %= p; tree[o << 1 | 1] %= p; lazy[o] = 0; } } void build(int o,int l,int r) { if(l == r) { tree[o] = a[rk[l]]; tree[o] %= p; return; } int mid = (l + r) >> 1; build(o << 1,l,mid); build(o << 1 | 1,mid + 1,r); tree[o] = tree[o << 1] + tree[o << 1 | 1]; tree[o] %= p; } void up_num(int o,int l,int r,int x,int y,int w) { if(l == x && r == y) { tree[o] += w * (l - r + 1); tree[o] %= p; lazy[o] += w; lazy[o] %= p; return; } push_down(o,l,r); int mid = (l + r) >> 1; if(mid < x) up_num(o << 1 | 1,mid + 1,r,x,y,w); else if(mid >= y) up_num(o << 1,l,mid,x,y,w); else { up_num(o << 1,l,mid,x,mid,w); up_num(o << 1 | 1,mid + 1,r,mid + 1,y,w); } tree[o] = tree[o << 1] + tree[o << 1 | 1]; tree[o] %= p; } int query(int o,int l,int r,int x,int y) { if(l == r && x == y) { return tree[o]; } push_down(o,l,r); int mid = (l + r) >> 1; if(mid >= y) return query(o << 1,l,mid,x,y); else if(mid < x) return query(o << 1 | 1,mid + 1,r,x,y); else { return (query(o << 1,l,mid,x,mid) + query(o << 1 | 1,mid + 1,r,mid + 1,y)) % p; } } int pathquery(int x,int y) { int ans = 0; while(top[x] != top[y]) { if(d[top[x]] < d[top[y]]) swap(x,y); ans += query(1,1,n,id[top[x]],id[x]); ans %= p; x = f[top[x]]; } if(d[x] > d[y]) swap(x,y); ans += query(1,1,n,id[x],id[y]); ans %= p; return ans; } void pathupdate(int x,int y,int c) { // int fx = top[x],fy = top[y]; while(top[x] != top[y]) { if(d[top[x]] < d[top[y]]) swap(x,y); up_num(1,1,n,id[top[x]],id[x],c); x = f[top[x]]; // update(id[x]) } if(d[x] > d[y]) swap(x,y); up_num(1,1,n,id[x],id[y],c); } int main() { read(n);read(m);read(r);read(p); duke(i,1,n) read(a[i]); duke(i,1,n - 1) { int x,y; read(x);read(y); add(x,y); add(y,x); } cnt = 0; dfs1(r,0,1); dfs2(r,r); cnt = 0; build(1,1,n); duke(i,1,m) { int op,x,y,z; read(op); if(op == 1) { read(x);read(y);read(z); pathupdate(x,y,z); } else if(op == 2) { read(x);read(y); printf("%d\n",pathquery(x,y)); } else if(op == 3) { read(x);read(z); // cout<<x<<endl; up_num(1,1,n,id[x],id[x] + siz[x] - 1,z); } else { read(x); printf("%d\n",query(1,1,n,id[x],id[x] + siz[x] - 1)); } } return 0; } /* 5 5 2 24 7 3 7 8 0 1 2 1 5 3 1 4 1 3 4 2 3 2 2 4 5 1 5 1 3 2 1 3 */
有人帮我debug吗？
标签：double code node ems read sizeof cti 负环 bug
原文地址：https://www.cnblogs.com/DukeLv/p/9693778.html
